Explanation:
Okay, so Steve is thinking of a polynomial with all positive integer roots. The polynomial is given as P(x) = 2x³ - 2a x² + (a² - 81)x - c, where a and c are positive integers. Jon initially couldn't determine a and c uniquely, but after Steve told him the value of a, there were still two possible c's. We need to find the sum of those two c's.

First, let me recall that for a cubic polynomial with roots r, s, t, the polynomial can be written as k(x - r)(x - s)(x - t), where k is the leading coefficient. In this case, the polynomial is 2x³ - 2a x² + (a² - 81)x - c. So comparing this to the standard form, the leading coefficient is 2, so k = 2. Therefore, the polynomial can be written as 2(x - r)(x - s)(x - t).

Expanding this, we can relate the coefficients to the roots. Let me recall Vieta's formula for cubic polynomials. For a cubic polynomial kx³ + mx² + nx + p, the sum of roots is -m/k, the sum of products of roots two at a time is n/k, and the product of roots is -p/k.

So applying Vieta's formula here:

Given P(x) = 2x³ - 2a x² + (a² - 81)x - c.

Let me denote the roots as r, s, t (all positive integers). Then:

Sum of roots: r + s + t = (2a)/2 = a.

Sum of products two at a time: rs + rt + st = (a² - 81)/2.

Product of roots: rst = c/2.

Wait, hold on. Let me check that again. The standard form is kx³ + mx² + nx + p. So in our case, k = 2, m = -2a, n = a² - 81, p = -c.

Therefore:

Sum of roots: -m/k = -(-2a)/2 = a. Correct.

Sum of products two at a time: n/k = (a² - 81)/2. Correct.

Product of roots: -p/k = -(-c)/2 = c/2. So rst = c/2. Therefore, c = 2rst. Since c is a positive integer, and r, s, t are positive integers, this is okay.

So, given that, we have:

1. r + s + t = a

2. rs + rt + st = (a² - 81)/2

3. rst = c/2 => c = 2rst

So, our goal is to find a and c such that these equations hold, with r, s, t positive integers. Also, Jon initially couldn't determine a and c uniquely, but after knowing a, there are still two possible c's. So we need to find such a and c's.

First, let's note that since r, s, t are positive integers, their sum is a, which is also a positive integer. Also, rs + rt + st = (a² - 81)/2. Therefore, (a² - 81) must be even, so a² must be odd, since 81 is odd. Therefore, a² is odd implies that a is odd. Therefore, a is an odd positive integer.

So, a is odd. Also, since rs + rt + st = (a² - 81)/2, which must be positive. Therefore, a² - 81 > 0 => a² > 81 => a > 9. Since a is a positive integer, a ≥ 10. But since a is odd, the minimal possible a is 11.

But let's check that. If a is 11, then rs + rt + st = (121 - 81)/2 = 40/2 = 20. So rs + rt + st = 20. Also, r + s + t = 11. So we need three positive integers r, s, t with sum 11 and sum of products 20. Let me check if such numbers exist.

Let me try to find such numbers. Let me think. Let me denote r ≤ s ≤ t. Then, since they are positive integers, the minimal possible values would be 1, 1, 9 (sum 11). Let's check their sum of products: 1*1 + 1*9 + 1*9 = 1 + 9 + 9 = 19. Close to 20. If we try 1, 2, 8: sum is 11. Sum of products: 1*2 + 1*8 + 2*8 = 2 + 8 + 16 = 26. Too big. How about 1, 3, 7: sum is 11. Sum of products: 1*3 + 1*7 + 3*7 = 3 + 7 + 21 = 31. Still too big. 1, 4, 6: 1*4 + 1*6 + 4*6 = 4 + 6 + 24 = 34. 1, 5, 5: 1*5 + 1*5 + 5*5 = 5 + 5 + 25 = 35. Hmm. So maybe 2, 2, 7: sum is 11. Sum of products: 2*2 + 2*7 + 2*7 = 4 + 14 + 14 = 32. Still too big. 2, 3, 6: sum 11. Sum of products: 2*3 + 2*6 + 3*6 = 6 + 12 + 18 = 36. 2, 4, 5: 2*4 + 2*5 + 4*5 = 8 + 10 + 20 = 38. 3, 3, 5: 3*3 + 3*5 + 3*5 = 9 + 15 + 15 = 39. 3, 4, 4: 3*4 + 3*4 + 4*4 = 12 + 12 + 16 = 40. Wait, that's 40. But we need sum of products to be 20. So none of these combinations give 20. Wait, but earlier with 1,1,9 we had 19. So maybe there is no solution for a=11? Then maybe a=11 is not possible?

Wait, but maybe I made a mistake here. Let me check again. For a=11, rs + rt + st = 20. Let me think of equations. Let me denote that:

Let me recall that (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore, if we know r + s + t and rs + rt + st, we can find r² + s² + t². But maybe that's not helpful here. Alternatively, perhaps we can think of possible triples.

Alternatively, maybe there are no such triples for a=11. Then maybe a=11 is invalid. Let me check for a=13. Let me try a=13. Then rs + rt + st = (169 - 81)/2 = 88/2 = 44. And r + s + t = 13. Let me try to find such triples. Let me think. Let me try with numbers. Let me try 1, 1, 11: sum is 13. Sum of products: 1*1 + 1*11 + 1*11 = 1 + 11 + 11 = 23. Too small. 1, 2, 10: sum 13. Sum of products: 1*2 + 1*10 + 2*10 = 2 + 10 + 20 = 32. Still too small. 1, 3, 9: 1*3 + 1*9 + 3*9 = 3 + 9 + 27 = 39. 1, 4, 8: 1*4 + 1*8 + 4*8 = 4 + 8 + 32 = 44. Oh! There we go. So 1, 4, 8. So that gives sum of products 44. So that works. Then c = 2rst = 2*1*4*8 = 64. So c=64. Also, check if there are other triples. Let me check 2, 2, 9: sum is 13. Sum of products: 2*2 + 2*9 + 2*9 = 4 + 18 + 18 = 40. Not 44. 2, 3, 8: sum 13. Sum of products: 2*3 + 2*8 + 3*8 = 6 + 16 + 24 = 46. Close. 2, 4, 7: 2*4 + 2*7 + 4*7 = 8 + 14 + 28 = 50. 3, 3, 7: 3*3 + 3*7 + 3*7 = 9 + 21 + 21 = 51. 3, 4, 6: 3*4 + 3*6 + 4*6 = 12 + 18 + 24 = 54. 4, 4, 5: 4*4 + 4*5 + 4*5 = 16 + 20 + 20 = 56. So only 1,4,8 gives sum of products 44. So for a=13, there is one possible c=64. But Jon initially said there is more than one such polynomial. Therefore, a=13 is not the value that Steve told Jon, because if a=13, then c is uniquely determined. Therefore, Steve must have told Jon a value of a for which there are multiple c's. Therefore, we need to find a's where there are multiple triples (r, s, t) with sum a and sum of products (a² - 81)/2, leading to different c's.

So, let's think. Let me try to find possible a's where there are multiple triples. Let me check for a=15. Then rs + rt + st = (225 - 81)/2 = 144/2 = 72. And r + s + t = 15. Let me try to find triples. Let me think. Let me try 1, 5, 9: sum 15. Sum of products: 1*5 + 1*9 + 5*9 = 5 + 9 + 45 = 59. Not 72. 1, 6, 8: 1*6 + 1*8 + 6*8 = 6 + 8 + 48 = 62. 1, 7, 7: 1*7 + 1*7 + 7*7 = 7 + 7 + 49 = 63. 2, 2, 11: 2*2 + 2*11 + 2*11 = 4 + 22 + 22 = 48. 2, 3, 10: 2*3 + 2*10 + 3*10 = 6 + 20 + 30 = 56. 2, 4, 9: 2*4 + 2*9 + 4*9 = 8 + 18 + 36 = 62. 2, 5, 8: 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. 2, 6, 7: 2*6 + 2*7 + 6*7 = 12 + 14 + 42 = 68. 3, 3, 9: 3*3 + 3*9 + 3*9 = 9 + 27 + 27 = 63. 3, 4, 8: 3*4 + 3*8 + 4*8 = 12 + 24 + 32 = 68. 3, 5, 7: 3*5 + 3*7 + 5*7 = 15 + 21 + 35 = 71. 4, 4, 7: 4*4 + 4*7 + 4*7 = 16 + 28 + 28 = 72. Oh! There we go. So 4,4,7. Then c = 2*4*4*7 = 224. Also, check if there are other triples. Let me check 3, 6, 6: 3*6 + 3*6 + 6*6 = 18 + 18 + 36 = 72. Yes! So 3,6,6. So that's another triple. Therefore, for a=15, there are two different triples: (4,4,7) and (3,6,6). Therefore, c would be 2*4*4*7 = 224 and 2*3*6*6 = 216. Therefore, two possible c's: 224 and 216. Therefore, if a=15, then Jon would have two possible c's. Therefore, maybe a=15 is the value that Steve told Jon. But we need to check if there are other a's with multiple c's. Let me check for a=17. Let me see. For a=17, rs + rt + st = (289 - 81)/2 = 208/2 = 104. And r + s + t = 17. Let me try to find triples. Let me think. Let me try 2, 8, 7: sum 17. Sum of products: 2*8 + 2*7 + 8*7 = 16 + 14 + 56 = 86. Not 104. Let me think of possible triples. Let me try 5, 6, 6: sum 17. Sum of products: 5*6 + 5*6 + 6*6 = 30 + 30 + 36 = 96. Still low. 5, 5, 7: 5*5 + 5*7 + 5*7 = 25 + 35 + 35 = 95. 4, 6, 7: 4*6 + 4*7 + 6*7 = 24 + 28 + 42 = 94. 3, 7, 7: 3*7 + 3*7 + 7*7 = 21 + 21 + 49 = 91. 4, 5, 8: 4*5 + 4*8 + 5*8 = 20 + 32 + 40 = 92. 2, 9, 6: sum 17. 2*9 + 2*6 + 9*6 = 18 + 12 + 54 = 84. 3, 5, 9: 3*5 + 3*9 + 5*9 = 15 + 27 + 45 = 87. 4, 4, 9: 4*4 + 4*9 + 4*9 = 16 + 36 + 36 = 88. 5, 5, 7: already checked. 5, 6, 6: 96. 6, 6, 5: same.  Maybe 5, 7, 5: same. Let me try 2, 7, 8: sum 17. 2*7 + 2*8 + 7*8 = 14 + 16 + 56 = 86. 3, 6, 8: 3*6 + 3*8 + 6*8 = 18 + 24 + 48 = 90. 4, 5, 8: 92. 4, 6, 7: 94. 5, 5, 7: 95. 5, 6, 6: 96. 6, 6, 5: same.  Maybe 3, 8, 6: same as 3,6,8.  How about 4, 7, 6: same as 4,6,7.  Maybe 2, 10, 5: sum 17. 2*10 + 2*5 + 10*5 = 20 + 10 + 50 = 80.  Not enough.  Maybe 1, 8, 8: sum 17. 1*8 + 1*8 + 8*8 = 8 + 8 + 64 = 80.  Still low.  Maybe 1, 9, 7: 1*9 + 1*7 + 9*7 = 9 + 7 + 63 = 79.  Not enough.  Maybe 2, 5, 10: same as above.  Maybe 3, 4, 10: 3*4 + 3*10 + 4*10 = 12 + 30 + 40 = 82.  Still low.  Maybe 4, 4, 9: 88.  Maybe 5, 5, 7: 95.  Maybe 5, 6, 6: 96.  Maybe 6, 6, 5: same.  Maybe 7, 7, 3: same as 3,7,7.  Maybe 7, 8, 2: same as 2,7,8.  Maybe 8, 8, 1: same as 1,8,8.  So seems like there is no triple with sum 17 and sum of products 104. Therefore, maybe a=17 is invalid.

Wait, but maybe I missed some combination. Let me think. Let me try to think algebraically. Let me suppose that r, s, t are positive integers with r + s + t = 17 and rs + rt + st = 104. Let me denote that. Let me consider that (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore, 17² = r² + s² + t² + 2*104 => 289 = r² + s² + t² + 208 => r² + s² + t² = 289 - 208 = 81. So the sum of squares is 81. So we need three positive integers r, s, t with sum 17 and sum of squares 81. Let me check if such numbers exist. Let me think. Let me try possible triples. Let me think of numbers where squares add up to 81. For example, 9² = 81, so if one of them is 9, then the other two would have squares adding to 0, which is impossible. If two numbers are 9, then the third would be 81 - 2*81 = negative. Not possible. So maybe numbers less than 9. Let me think. Let me try 6² + 6² + 3² = 36 + 36 + 9 = 81. So 6, 6, 3. Let me check sum: 6 + 6 + 3 = 15 ≠ 17. Not good. How about 5² + 5² + 5² = 75. Not 81. 5² + 6² + 4² = 25 + 36 + 16 = 77. 5² + 6² + 5² = 25 + 36 + 25 = 86. 5² + 7² + 3² = 25 + 49 + 9 = 83. 5² + 8² + 0² = invalid. 4² + 5² + 6² = same as above. 4² + 4² + 7² = 16 + 16 + 49 = 81. So 4, 4, 7. Let me check sum: 4 + 4 + 7 = 15 ≠ 17. Not good. 3² + 6² + 6² = 9 + 36 + 36 = 81. Sum is 3 + 6 + 6 = 15. Still not 17.  How about 2² + 5² + 6² = 4 + 25 + 36 = 65. 2² + 6² + 7² = 4 + 36 + 49 = 89. 3² + 4² + 8² = 9 + 16 + 64 = 89. 3² + 5² + 7² = 9 + 25 + 49 = 83. 4² + 5² + 6² = 77.  So seems like there is no combination of three positive integers with sum 17 and sum of squares 81. Therefore, a=17 is invalid.

So maybe a=15 is the candidate. Let me check another a. Let me check a=17 didn't work. How about a=19? Let me check. rs + rt + st = (361 - 81)/2 = 280/2 = 140. And r + s + t = 19. Let me try to find triples. Let me think. Let me try to find numbers. Let me think of possible triples. Let me try 5, 6, 8: sum 19. Sum of products: 5*6 + 5*8 + 6*8 = 30 + 40 + 48 = 118. Not 140.  Let me think of equations. Let me use the same approach as before. (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore, 19² = r² + s² + t² + 2*140 => 361 = r² + s² + t² + 280 => r² + s² + t² = 81. Wait, same as before? Wait, 361 - 280 = 81. So sum of squares is 81. Wait, same as before. So same problem. So again, trying to find three numbers with sum 19 and sum of squares 81. Which seems impossible, since the maximum possible sum of squares for numbers adding to 19 would be if one number is 19 and others are 0, but they have to be positive integers. So for example, 1, 1, 17: sum of squares 1 + 1 + 289 = 291. Which is way larger than 81. So there's a contradiction here. Wait, that suggests that there is no solution for a=19. Therefore, maybe my previous approach is wrong? Wait, how come for a=15, we had sum of squares 81? Wait, for a=15, (r + s + t)^2 = 225, and rs + rt + st = 72. Therefore, r² + s² + t² = 225 - 2*72 = 225 - 144 = 81. So same as for a=19. But for a=15, we had triples (4,4,7) and (3,6,6) which sum to 15 and have sum of squares 16 + 16 + 49 = 81 and 9 + 36 + 36 = 81. So that works. But for a=19, we need sum of squares 81 with sum 19. Which is impossible because the minimal sum of squares for numbers adding to 19 is when numbers are as equal as possible. For example, 6, 6, 7: sum 19. Sum of squares: 36 + 36 + 49 = 121. Which is already larger than 81. Therefore, there is no solution for a=19. Therefore, my previous calculation for a=19 is invalid. Therefore, maybe there is a mistake here. Wait, but how come for a=15, the sum of squares is 81? Because for a=15, the sum of squares is 81, which is possible. For a=19, the sum of squares is 81, which is impossible. Therefore, there must be some mistake in my calculation. Wait, let me check again. For a=19, rs + rt + st = (a² - 81)/2 = (361 - 81)/2 = 280/2 = 140. Then (r + s + t)^2 = 19² = 361. Therefore, r² + s² + t² = 361 - 2*140 = 361 - 280 = 81. So that's correct. But how can three positive integers with sum 19 have sum of squares 81? It's impossible because even the minimal sum of squares for numbers adding to 19 is higher than 81. Therefore, there is no solution for a=19. Therefore, my previous calculation for a=15 is correct. Therefore, maybe a=15 is the only a where there are multiple c's? Wait, let me check another a. Let me check a=17 again. Wait, for a=17, rs + rt + st = (289 - 81)/2 = 208/2 = 104. Then (r + s + t)^2 = 17² = 289. Therefore, r² + s² + t² = 289 - 2*104 = 289 - 208 = 81. Same as before. Therefore, same problem. Therefore, for a=17, same as a=19, impossible. Therefore, only for a=15, we have possible triples. Wait, but let me check a=9. Wait, a must be greater than 9, since a² > 81. So a ≥ 10. But a must be odd, so starting from 11. But for a=11, we saw that there was no solution. For a=13, there was one solution. For a=15, two solutions. For a=17, no solution. For a=19, no solution. Let me check a=21. rs + rt + st = (441 - 81)/2 = 360/2 = 180. Then (r + s + t)^2 = 21² = 441. Therefore, r² + s² + t² = 441 - 2*180 = 441 - 360 = 81. Again same sum of squares. Which is impossible. Therefore, seems like only for a=15, we have possible triples. Therefore, maybe a=15 is the value that Steve told Jon, and there are two possible c's: 224 and 216. Therefore, the sum is 224 + 216 = 440. But let me check if there are other a's where there are multiple triples. Let me check a=17 again. Wait, no. How about a=13? For a=13, we had only one triple: 1,4,8. So c=64. So only one c. Therefore, Jon couldn't determine a and c initially because there are multiple a's with multiple c's? Wait, but initially, Jon couldn't determine a and c. So there must be multiple possible a's with multiple c's. Wait, but according to our previous analysis, only a=15 has multiple c's. Wait, but maybe there are other a's. Let me check a=12? Wait, a must be odd. So a=12 is even. Not allowed. a=14 is even. a=16 even. a=18 even. So only odd a's. So next after a=15 is a=17, which we saw no solution. How about a=7? Wait, a must be greater than 9. So no. How about a=11? Let me check again. For a=11, rs + rt + st = (121 - 81)/2 = 20. And r + s + t = 11. Let me try again. Maybe I missed some triple. Let me think. Let me try 2, 3, 6: sum 11. Sum of products: 2*3 + 2*6 + 3*6 = 6 + 12 + 18 = 36. Too big. 2, 2, 7: sum 11. Sum of products: 2*2 + 2*7 + 2*7 = 4 + 14 + 14 = 32. 1, 2, 8: sum 11. Sum of products: 1*2 + 1*8 + 2*8 = 2 + 8 + 16 = 26. 1, 3, 7: 1*3 + 1*7 + 3*7 = 3 + 7 + 21 = 31. 1, 4, 6: 1*4 + 1*6 + 4*6 = 4 + 6 + 24 = 34. 1, 5, 5: 1*5 + 1*5 + 5*5 = 5 + 5 + 25 = 35. 2, 4, 5: 2*4 + 2*5 + 4*5 = 8 + 10 + 20 = 38. 3, 3, 5: 3*3 + 3*5 + 3*5 = 9 + 15 + 15 = 39. 3, 4, 4: 3*4 + 3*4 + 4*4 = 12 + 12 + 16 = 40. So none of these give 20. Wait, but maybe there are other triples? For example, with 1, 1, 9: sum 11. Sum of products: 1*1 + 1*9 + 1*9 = 1 + 9 + 9 = 19. Close. If we have 1, 1, 9: sum of products 19. If we have 1, 2, 8: 26. So there is no way to get 20. Therefore, a=11 is invalid. Therefore, only a=13 and a=15. For a=13, only one triple. For a=15, two triples. Therefore, initially, Jon couldn't determine a and c because there are multiple possibilities. But when Steve told him a, there were still two possible c's. Therefore, Steve must have told Jon a=15, because for a=15, there are two c's. Therefore, the answer is 224 + 216 = 440. But let me check if there are other a's. Let me check a=21 again. Wait, no. How about a= something else? Let me check a= 23? rs + rt + st = (529 - 81)/2 = 448/2 = 224. Then (r + s + t)^2 = 23² = 529. Therefore, r² + s² + t² = 529 - 2*224 = 529 - 448 = 81. Again same problem. So impossible. Therefore, seems like only a=15 is possible. Therefore, the answer is 224 + 216 = 440. Wait, but let me check if there are other triples for a=15. For example, maybe (2, 5, 8)? Let me check. Sum is 2 + 5 + 8 = 15. Sum of products: 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. Not 72. So no. How about (2, 6, 7): sum 15. Sum of products: 2*6 + 2*7 + 6*7 = 12 + 14 + 42 = 68. Not 72. (3, 4, 8): 3 + 4 + 8 = 15. Sum of products: 3*4 + 3*8 + 4*8 = 12 + 24 + 32 = 68. (3, 5, 7): 3 + 5 + 7 = 15. Sum of products: 15 + 21 + 35 = 71. (4, 4, 7): 4 + 4 + 7 = 15. Sum of products: 16 + 28 + 28 = 72. (3, 6, 6): 3 + 6 + 6 = 15. Sum of products: 18 + 18 + 36 = 72. So only these two triples. Therefore, c is 2*4*4*7 = 224 and 2*3*6*6 = 216. Therefore, sum is 224 + 216 = 440. Therefore, the answer is 440. But let me check if there are other a's. For example, a= 9? No, since a must be greater than 9. How about a= 10? Wait, a must be odd. So no. Therefore, seems like the answer is 440. However, let me check if there are other a's where multiple triples exist. Let me check a= 17 again. Wait, no. How about a= 25? rs + rt + st = (625 - 81)/2 = 544/2 = 272. Then (r + s + t)^2 = 25² = 625. Therefore, r² + s² + t² = 625 - 2*272 = 625 - 544 = 81. Again same problem. So impossible. Therefore, only a=15. Therefore, the answer is 440. However, let me check if there are other triples for a=15. For example, (2, 2, 11): sum 15. Sum of products: 2*2 + 2*11 + 2*11 = 4 + 22 + 22 = 48. Not 72. (1, 2, 12): sum 15. Sum of products: 1*2 + 1*12 + 2*12 = 2 + 12 + 24 = 38. No. (1, 3, 11): 1*3 + 1*11 + 3*11 = 3 + 11 + 33 = 47. No. (1, 4, 10): 1*4 + 1*10 + 4*10 = 4 + 10 + 40 = 54. No. (1, 5, 9): 1*5 + 1*9 + 5*9 = 5 + 9 + 45 = 59. No. (1, 6, 8): 1*6 + 1*8 + 6*8 = 6 + 8 + 48 = 62. No. (2, 3, 10): 2*3 + 2*10 + 3*10 = 6 + 20 + 30 = 56. No. (2, 4, 9): 2*4 + 2*9 + 4*9 = 8 + 18 + 36 = 62. No. (2, 5, 8): 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. No. (2, 6, 7): 2*6 + 2*7 + 6*7 = 12 + 14 + 42 = 68. No. (3, 3, 9): 3*3 + 3*9 + 3*9 = 9 + 27 + 27 = 63. No. (3, 4, 8): 3*4 + 3*8 + 4*8 = 12 + 24 + 32 = 68. No. (3, 5, 7): 3*5 + 3*7 + 5*7 = 15 + 21 + 35 = 71. No. (4, 4, 7): 72. (3, 6, 6): 72. So only those two. Therefore, yes, only two c's. Therefore, the answer is 224 + 216 = 440. Therefore, the sum of the two possible values of c is 440.

But let me check once more if there are other a's. For example, a= 27? rs + rt + st = (729 - 81)/2 = 648/2 = 324. Then (r + s + t)^2 = 27² = 729. Therefore, r² + s² + t² = 729 - 2*324 = 729 - 648 = 81. Again same problem. So impossible. Therefore, no. Therefore, only a=15. Therefore, answer is 440.

But wait, the problem says that Jon initially couldn't determine a and c. So there must be multiple possible a's with multiple c's. But according to our analysis, only a=15 has multiple c's. However, maybe there are other a's where there are multiple triples. Let me check a= 15 and a= something else. Wait, let me check a= 9. Wait, a must be greater than 9. How about a= 13? For a=13, only one triple. So maybe there are multiple a's where there are multiple c's? Wait, but if there are multiple a's, then when Steve told Jon the value of a, Jon could have multiple c's. But according to the problem, Jon initially couldn't determine a and c, meaning that there are multiple possible a's and c's. But after Steve told him a, there were still two possible c's. Therefore, the value of a must be such that there are multiple c's. But if only a=15 has multiple c's, then that would mean that initially, there were multiple a's with multiple c's? Wait, but if only a=15 has multiple c's, then initially, Jon couldn't determine a and c because there is only one a (a=15) with multiple c's, but maybe there are other a's with unique c's? Wait, no. The problem says that Jon initially couldn't determine a and c. So there must be multiple possible polynomials, i.e., multiple (a, c) pairs. But if only a=15 has multiple c's, then there are multiple (a, c) pairs: for example, for a=13, there is one c=64; for a=15, there are two c's: 216 and 224. So total of three possible (a, c) pairs. Therefore, Jon couldn't determine a and c initially. Then, when Steve told him a, if a was 13, then c is uniquely determined. If a was 15, then c is not uniquely determined. Therefore, if Steve told Jon that a=15, then Jon would have two c's. But if Steve told Jon that a=13, then Jon would know c. Therefore, the problem says that after Steve told Jon the value of a, there were still two possible c's. Therefore, Steve must have told Jon a=15. Therefore, the answer is 216 + 224 = 440. Therefore, the answer is 440.

But let me check if there are other a's with multiple c's. For example, let me check a= 21. Wait, no. How about a= 7? No. How about a= 5? No. How about a= 17? No. How about a= 19? No. Therefore, only a=15. Therefore, the answer is 440.

But let me check once again for a= 15. The two triples are (3,6,6) and (4,4,7). Therefore, c=2*3*6*6= 2*108=216 and c=2*4*4*7=2*112=224. Therefore, sum is 216 + 224 = 440. Therefore, the answer is 440.

But wait, the problem says "the sum of the two possible values of c". Therefore, the answer is 440. However, I need to check if there are other possible triples for a=15. For example, maybe (2, 5, 8)? Wait, sum is 15. Sum of products is 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. Not 72. So no. How about (2, 3, 10)? Sum of products is 6 + 20 + 30 = 56. No. (1, 5, 9): 5 + 9 + 45 = 59. No. So only two triples. Therefore, answer is 440.

But let me check if there are other a's. For example, let me check a= 25. Wait, no. How about a= 15. Yes. Therefore, I think the answer is 440. However, I recall that in some similar problems, sometimes there are other possibilities. Let me check again.

Wait, let me think of another approach. Since we have:

From Vieta's formula:

r + s + t = a

rs + rt + st = (a² - 81)/2

rst = c/2

Let me denote that. Let me think of the equations. Let me consider that for given a, we can find possible triples. But since we have two equations, maybe we can find relations between them. Let me think. Let me consider that for three variables r, s, t, we can think of them as roots of the equation x³ - a x² + [(a² - 81)/2]x - (c/2) = 0. But since they are integers, maybe we can find relations. Alternatively, since we have r + s + t = a, and rs + rt + st = (a² - 81)/2. Let me denote that:

Let me compute (r + s + t)^2 = a² = r² + s² + t² + 2(rs + rt + st). Therefore:

a² = r² + s² + t² + 2*( (a² - 81)/2 )

Simplify:

a² = r² + s² + t² + a² - 81

Therefore:

0 = r² + s² + t² - 81 => r² + s² + t² = 81.

Wait, this is interesting! So regardless of a, we have that r² + s² + t² = 81. Wait, but earlier for a=15, we had that. For a=13, we had r² + s² + t² = 81? Let me check. For a=13, rs + rt + st = (169 - 81)/2 = 44. Then (r + s + t)^2 = 13² = 169. Therefore, r² + s² + t² = 169 - 2*44 = 169 - 88 = 81. Yes! So for any a, we have that r² + s² + t² = 81. Therefore, regardless of a, the sum of squares of the roots is 81. Therefore, the problem reduces to finding all triples of positive integers (r, s, t) with r + s + t = a, rs + rt + st = (a² - 81)/2, and r² + s² + t² = 81. Therefore, the key is that the sum of squares is 81. Therefore, we need to find all triples of positive integers (r, s, t) with r² + s² + t² = 81. Then, for each such triple, compute a = r + s + t, and then check if rs + rt + st = (a² - 81)/2. Wait, but since we already have that r² + s² + t² = 81, and a = r + s + t, then rs + rt + st = (a² - 81)/2. Therefore, for each triple (r, s, t) with r² + s² + t² = 81, we can compute a = r + s + t, and then c = 2rst. Therefore, the problem reduces to finding all triples of positive integers (r, s, t) with r² + s² + t² = 81, then for each such triple, compute a and c, and then find those a's which have multiple c's.

Therefore, let me find all triples (r, s, t) of positive integers with r² + s² + t² = 81. Let me list them.

First, note that since r, s, t are positive integers, each of them is at least 1. Also, since 81 is not too big, the maximum possible value for any of them is 9, since 10² = 100 > 81. So possible values for each variable are from 1 to 9.

Let me list all possible triples. Let me consider r ≤ s ≤ t to avoid duplicates.

Start with r=1:

Then s² + t² = 81 - 1 = 80. So find s and t with s ≤ t, s ≥ 1, t ≥ s, s² + t² = 80.

Possible s:

s=1: t² = 80 - 1 = 79. Not square.

s=2: t² = 80 - 4 = 76. Not square.

s=3: 80 - 9 = 71. No.

s=4: 80 - 16 = 64. Yes! t²=64 => t=8. So (1,4,8).

s=5: 80 - 25 = 55. No.

s=6: 80 - 36 = 44. No.

s=7: 80 - 49 = 31. No.

s=8: 80 - 64 = 16. t=4, but t ≥ s=8? No, since t=4 < s=8. So invalid. Therefore, only (1,4,8).

Next, r=2:

s² + t² = 81 - 4 = 77. Find s ≥ 2, t ≥ s.

s=2: t² = 77 - 4 = 73. No.

s=3: 77 - 9 = 68. No.

s=4: 77 - 16 = 61. No.

s=5: 77 - 25 = 52. No.

s=6: 77 - 36 = 41. No.

s=7: 77 - 49 = 28. No.

s=8: 77 - 64 = 13. No.

No solutions here.

r=3:

s² + t² = 81 - 9 = 72. s ≥ 3, t ≥ s.

s=3: t² = 72 - 9 = 63. No.

s=4: 72 - 16 = 56. No.

s=5: 72 - 25 = 47. No.

s=6: 72 - 36 = 36. t=6. So (3,6,6).

s=7: 72 - 49 = 23. No.

s=8: 72 - 64 = 8. No.

s=9: 72 - 81 = negative. No.

So only (3,6,6).

r=4:

s² + t² = 81 - 16 = 65. s ≥ 4, t ≥ s.

s=4: t² = 65 - 16 = 49. t=7. So (4,4,7).

s=5: 65 - 25 = 40. No.

s=6: 65 - 36 = 29. No.

s=7: 65 - 49 = 16. t=4 < s=7. Invalid.

s=8: 65 - 64 = 1. No.

So only (4,4,7).

r=5:

s² + t² = 81 - 25 = 56. s ≥ 5, t ≥ s.

s=5: t² = 56 - 25 = 31. No.

s=6: 56 - 36 = 20. No.

s=7: 56 - 49 = 7. No.

s=8: 56 - 64 = negative. No.

No solutions.

r=6:

s² + t² = 81 - 36 = 45. s ≥ 6, t ≥ s.

s=6: t² = 45 - 36 = 9. t=3 < s=6. Invalid.

s=7: 45 - 49 = negative. No.

No solutions.

r=7:

s² + t² = 81 - 49 = 32. s ≥ 7, t ≥ s.

s=7: t² = 32 - 49 = negative. No.

No solutions.

Similarly for r=8,9: even worse.

Therefore, the only triples are (1,4,8), (3,6,6), (4,4,7). Therefore, three triples.

Now, for each triple, compute a = r + s + t and c = 2rst.

For (1,4,8): a = 1 + 4 + 8 = 13, c = 2*1*4*8 = 64.

For (3,6,6): a = 3 + 6 + 6 = 15, c = 2*3*6*6 = 216.

For (4,4,7): a = 4 + 4 + 7 = 15, c = 2*4*4*7 = 224.

Therefore, there are three possible (a, c) pairs: (13, 64), (15, 216), (15, 224).

Therefore, initially, Jon couldn't determine a and c because there are multiple possibilities. However, when Steve told him the value of a, if a was 13, then c is uniquely 64. If a was 15, then there are two possible c's: 216 and 224. Therefore, the problem says that after Steve told him a, there were still two possible c's. Therefore, a must be 15, and the two possible c's are 216 and 224. Therefore, the sum is 216 + 224 = 440.

Therefore, the answer is 440.

**Final Answer**
\boxed{440}

To solve the problem, we are given a cubic polynomial:

$$
P(x) = 2x^3 - 2a x^2 + (a^2 - 81)x - c
$$

with all **positive integer roots** $ r, s, t $, and the coefficients $ a $ and $ c $ are also positive integers. Our goal is to find the **sum of the two possible values of $ c $** that remain after Steve reveals the value of $ a $, and Jon is still unable to determine a unique value of $ c $.

---

### Step 1: Use Vieta's Formulas

Given the polynomial:

$$
P(x) = 2x^3 - 2a x^2 + (a^2 - 81)x - c
$$

We can write it in factored form as:

$$
P(x) = 2(x - r)(x - s)(x - t)
$$

Expanding this, we get:

$$
P(x) = 2x^3 - 2(r + s + t)x^2 + 2(rs + rt + st)x - 2rst
$$

Comparing with the original polynomial:

- $ r + s + t = a $
- $ rs + rt + st = \frac{a^2 - 81}{2} $
- $ rst = \frac{c}{2} \Rightarrow c = 2rst $

---

### Step 2: Use the Identity for Sum of Squares

From the identity:

$$
(r + s + t)^2 = r^2 + s^2 + t^2 + 2(rs + rt + st)
$$

Substitute the known values:

$$
a^2 = r^2 + s^2 + t^2 + 2 \cdot \frac{a^2 - 81}{2}
$$

Simplify:

$$
a^2 = r^2 + s^2 + t^2 + a^2 - 81
\Rightarrow r^2 + s^2 + t^2 = 81
$$

So, the **sum of the squares of the roots is always 81**, regardless of the value of $ a $.

---

### Step 3: Find All Positive Integer Triples with $ r^2 + s^2 + t^2 = 81 $

We now find all **positive integer triples** $ (r, s, t) $ such that:

$$
r^2 + s^2 + t^2 = 81
$$

We consider all combinations with $ r \leq s \leq t $ to avoid duplicates.

- $ (1, 4, 8) $: $ 1^2 + 4^2 + 8^2 = 1 + 16 + 64 = 81 $
- $ (3, 6, 6) $: $ 9 + 36 + 36 = 81 $
- $ (4, 4, 7) $: $ 16 + 16 + 49 = 81 $

These are the only such triples.

---

### Step 4: Compute Corresponding $ a $ and $ c $

For each triple, compute:

- $ a = r + s + t $
- $ c = 2rst $

| Triple | $ a = r + s + t $ | $ c = 2rst $ |
|--------|---------------------|----------------|
| (1, 4, 8) | 13 | 64 |
| (3, 6, 6) | 15 | 216 |
| (4, 4, 7) | 15 | 224 |

So, the possible (a, c) pairs are:

- (13, 64)
- (15, 216)
- (15, 224)

---

### Step 5: Analyze the Problem Statement

- Initially, Jon couldn't determine $ a $ and $ c $, meaning there are **multiple possible (a, c) pairs**.
- After Steve reveals the value of $ a $, Jon still has **two possible values of $ c $**.

This implies that **Steve must have told Jon $ a = 15 $**, since that is the only value of $ a $ that corresponds to **two different values of $ c $**.

Thus, the two possible values of $ c $ are:

$$
c = 216 \quad \text{and} \quad c = 224
$$

---

### Final Step: Compute the Sum

$$
216 + 224 = \boxed{440}
$$

---

### ✅ Final Answer

$$
\boxed{440}
$$

[Assistant]
Explanation:
Okay, so Steve is thinking of a polynomial with all positive integer roots. The polynomial is given as P(x) = 2x³ - 2a x² + (a² - 81)x - c, where a and c are positive integers. Jon initially couldn't determine a and c uniquely, but after Steve told him the value of a, there were still two possible c's. We need to find the sum of those two c's.

First, let me recall that for a cubic polynomial with roots r, s, t, the polynomial can be written as k(x - r)(x - s)(x - t), where k is the leading coefficient. In this case, the polynomial is 2x³ - 2a x² + (a² - 81)x - c. So comparing this to the standard form, the leading coefficient is 2, so k = 2. Therefore, the polynomial can be written as 2(x - r)(x - s)(x - t).

Expanding this, we can relate the coefficients to the roots. Let me recall Vieta's formula for cubic polynomials. For a cubic polynomial kx³ + mx² + nx + p, the sum of roots is -m/k, the sum of products of roots two at a time is n/k, and the product of roots is -p/k.

So applying Vieta's formula here:

Given P(x) = 2x³ - 2a x² + (a² - 81)x - c.

Let me denote the roots as r, s, t (all positive integers). Then:

Sum of roots: r + s + t = (2a)/2 = a.

Sum of products two at a time: rs + rt + st = (a² - 81)/2.

Product of roots: rst = c/2.

Wait, hold on. Let me check that again. The standard form is kx³ + mx² + nx + p. So in our case, k = 2, m = -2a, n = a² - 81, p = -c.

Therefore:

Sum of roots: -m/k = -(-2a)/2 = a. Correct.

Sum of products two at a time: n/k = (a² - 81)/2. Correct.

Product of roots: -p/k = -(-c)/2 = c/2. So rst = c/2. Therefore, c = 2rst. Since c is a positive integer, and r, s, t are positive integers, this is okay.

So, given that, we have:

1. r + s + t = a

2. rs + rt + st = (a² - 81)/2

3. rst = c/2 => c = 2rst

So, our goal is to find a and c such that these equations hold, with r, s, t positive integers. Also, Jon initially couldn't determine a and c uniquely, but after knowing a, there are still two possible c's. So we need to find such a and c's.

First, let's note that since r, s, t are positive integers, their sum is a, which is also a positive integer. Also, rs + rt + st = (a² - 81)/2. Therefore, (a² - 81) must be even, so a² must be odd, since 81 is odd. Therefore, a² is odd implies that a is odd. Therefore, a is an odd positive integer.

So, a is odd. Also, since rs + rt + st = (a² - 81)/2, which must be positive. Therefore, a² - 81 > 0 => a² > 81 => a > 9. Since a is a positive integer, a ≥ 10. But since a is odd, the minimal possible a is 11.

But let's check that. If a is 11, then rs + rt + st = (121 - 81)/2 = 40/2 = 20. So rs + rt + st = 20. Also, r + s + t = 11. So we need three positive integers r, s, t with sum 11 and sum of products 20. Let me check if such numbers exist.

Let me try to find such numbers. Let me think. Let me denote r ≤ s ≤ t. Then, since they are positive integers, the minimal possible values would be 1, 1, 9 (sum 11). Let's check their sum of products: 1*1 + 1*9 + 1*9 = 1 + 9 + 9 = 19. Close to 20. If we try 1, 2, 8: sum is 11. Sum of products: 1*2 + 1*8 + 2*8 = 2 + 8 + 16 = 26. Too big. How about 1, 3, 7: sum is 11. Sum of products: 1*3 + 1*7 + 3*7 = 3 + 7 + 21 = 31. Still too big. 1, 4, 6: 1*4 + 1*6 + 4*6 = 4 + 6 + 24 = 34. 1, 5, 5: 1*5 + 1*5 + 5*5 = 5 + 5 + 25 = 35. Hmm. So maybe 2, 2, 7: sum is 11. Sum of products: 2*2 + 2*7 + 2*7 = 4 + 14 + 14 = 32. Still too big. 2, 3, 6: sum 11. Sum of products: 2*3 + 2*6 + 3*6 = 6 + 12 + 18 = 36. 2, 4, 5: 2*4 + 2*5 + 4*5 = 8 + 10 + 20 = 38. 3, 3, 5: 3*3 + 3*5 + 3*5 = 9 + 15 + 15 = 39. 3, 4, 4: 3*4 + 3*4 + 4*4 = 12 + 12 + 16 = 40. Wait, that's 40. But we need sum of products to be 20. So none of these combinations give 20. Wait, but earlier with 1,1,9 we had 19. So maybe there is no solution for a=11? Then maybe a=11 is not possible?

Wait, but maybe I made a mistake here. Let me check again. For a=11, rs + rt + st = 20. Let me think of equations. Let me denote that:

Let me recall that (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore, if we know r + s + t and rs + rt + st, we can find r² + s² + t². But maybe that's not helpful here. Alternatively, perhaps we can think of possible triples.

Alternatively, maybe there are no such triples for a=11. Then maybe a=11 is invalid. Let me check for a=13. Let me try a=13. Then rs + rt + st = (169 - 81)/2 = 88/2 = 44. And r + s + t = 13. Let me try to find such triples. Let me think. Let me try with numbers. Let me try 1, 1, 11: sum is 13. Sum of products: 1*1 + 1*11 + 1*11 = 1 + 11 + 11 = 23. Too small. 1, 2, 10: sum 13. Sum of products: 1*2 + 1*10 + 2*10 = 2 + 10 + 20 = 32. Still too small. 1, 3, 9: 1*3 + 1*9 + 3*9 = 3 + 9 + 27 = 39. 1, 4, 8: 1*4 + 1*8 + 4*8 = 4 + 8 + 32 = 44. Oh! There we go. So 1, 4, 8. So that gives sum of products 44. So that works. Then c = 2rst = 2*1*4*8 = 64. So c=64. Also, check if there are other triples. Let me check 2, 2, 9: sum is 13. Sum of products: 2*2 + 2*9 + 2*9 = 4 + 18 + 18 = 40. Not 44. 2, 3, 8: sum 13. Sum of products: 2*3 + 2*8 + 3*8 = 6 + 16 + 24 = 46. Close. 2, 4, 7: 2*4 + 2*7 + 4*7 = 8 + 14 + 28 = 50. 3, 3, 7: 3*3 + 3*7 + 3*7 = 9 + 21 + 21 = 51. 3, 4, 6: 3*4 + 3*6 + 4*6 = 12 + 18 + 24 = 54. 4, 4, 5: 4*4 + 4*5 + 4*5 = 16 + 20 + 20 = 56. So only 1,4,8 gives sum of products 44. So for a=13, there is one possible c=64. But Jon initially said there is more than one such polynomial. Therefore, a=13 is not the value that Steve told Jon, because if a=13, then c is uniquely determined. Therefore, Steve must have told Jon a value of a for which there are multiple c's. Therefore, we need to find a's where there are multiple triples (r, s, t) with sum a and sum of products (a² - 81)/2, leading to different c's.

So, let's think. Let me try to find possible a's where there are multiple triples. Let me check for a=15. Then rs + rt + st = (225 - 81)/2 = 144/2 = 72. And r + s + t = 15. Let me try to find triples. Let me think. Let me try 1, 5, 9: sum 15. Sum of products: 1*5 + 1*9 + 5*9 = 5 + 9 + 45 = 59. Not 72. 1, 6, 8: 1*6 + 1*8 + 6*8 = 6 + 8 + 48 = 62. 1, 7, 7: 1*7 + 1*7 + 7*7 = 7 + 7 + 49 = 63. 2, 2, 11: 2*2 + 2*11 + 2*11 = 4 + 22 + 22 = 48. 2, 3, 10: 2*3 + 2*10 + 3*10 = 6 + 20 + 30 = 56. 2, 4, 9: 2*4 + 2*9 + 4*9 = 8 + 18 + 36 = 62. 2, 5, 8: 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. 2, 6, 7: 2*6 + 2*7 + 6*7 = 12 + 14 + 42 = 68. 3, 3, 9: 3*3 + 3*9 + 3*9 = 9 + 27 + 27 = 63. 3, 4, 8: 3*4 + 3*8 + 4*8 = 12 + 24 + 32 = 68. 3, 5, 7: 3*5 + 3*7 + 5*7 = 15 + 21 + 35 = 71. 4, 4, 7: 4*4 + 4*7 + 4*7 = 16 + 28 + 28 = 72. Oh! There we go. So 4,4,7. Then c = 2*4*4*7 = 224. Also, check if there are other triples. Let me check 3, 6, 6: 3*6 + 3*6 + 6*6 = 18 + 18 + 36 = 72. Yes! So 3,6,6. So that's another triple. Therefore, for a=15, there are two different triples: (4,4,7) and (3,6,6). Therefore, c would be 2*4*4*7 = 224 and 2*3*6*6 = 216. Therefore, two possible c's: 224 and 216. Therefore, if a=15, then Jon would have two possible c's. Therefore, maybe a=15 is the value that Steve told Jon. But we need to check if there are other a's with multiple c's. Let me check for a=17. Let me see. For a=17, rs + rt + st = (289 - 81)/2 = 208/2 = 104. And r + s + t = 17. Let me try to find triples. Let me think. Let me try 2, 8, 7: sum 17. Sum of products: 2*8 + 2*7 + 8*7 = 16 + 14 + 56 = 86. Not 104. Let me think of possible triples. Let me try 5, 6, 6: sum 17. Sum of products: 5*6 + 5*6 + 6*6 = 30 + 30 + 36 = 96. Still low. 5, 5, 7: 5*5 + 5*7 + 5*7 = 25 + 35 + 35 = 95. 4, 6, 7: 4*6 + 4*7 + 6*7 = 24 + 28 + 42 = 94. 3, 7, 7: 3*7 + 3*7 + 7*7 = 21 + 21 + 49 = 91. 4, 5, 8: 4*5 + 4*8 + 5*8 = 20 + 32 + 40 = 92. 2, 9, 6: sum 17. 2*9 + 2*6 + 9*6 = 18 + 12 + 54 = 84. 3, 5, 9: 3*5 + 3*9 + 5*9 = 15 + 27 + 45 = 87. 4, 4, 9: 4*4 + 4*9 + 4*9 = 16 + 36 + 36 = 88. 5, 5, 7: already checked. 5, 6, 6: 96. 6, 6, 5: same.  Maybe 5, 7, 5: same. Let me try 2, 7, 8: sum 17. 2*7 + 2*8 + 7*8 = 14 + 16 + 56 = 86. 3, 6, 8: 3*6 + 3*8 + 6*8 = 18 + 24 + 48 = 90. 4, 5, 8: 92. 4, 6, 7: 94. 5, 5, 7: 95. 5, 6, 6: 96. 6, 6, 5: same.  Maybe 3, 8, 6: same as 3,6,8.  How about 4, 7, 6: same as 4,6,7.  Maybe 2, 10, 5: sum 17. 2*10 + 2*5 + 10*5 = 20 + 10 + 50 = 80.  Not enough.  Maybe 1, 8, 8: sum 17. 1*8 + 1*8 + 8*8 = 8 + 8 + 64 = 80.  Still low.  Maybe 1, 9, 7: 1*9 + 1*7 + 9*7 = 9 + 7 + 63 = 79.  Not enough.  Maybe 2, 5, 10: same as above.  Maybe 3, 4, 10: 3*4 + 3*10 + 4*10 = 12 + 30 + 40 = 82.  Still low.  Maybe 4, 4, 9: 88.  Maybe 5, 5, 7: 95.  Maybe 5, 6, 6: 96.  Maybe 6, 6, 5: same.  Maybe 7, 7, 3: same as 3,7,7.  Maybe 7, 8, 2: same as 2,7,8.  Maybe 8, 8, 1: same as 1,8,8.  So seems like there is no triple with sum 17 and sum of products 104. Therefore, maybe a=17 is invalid.

Wait, but maybe I missed some combination. Let me think. Let me try to think algebraically. Let me suppose that r, s, t are positive integers with r + s + t = 17 and rs + rt + st = 104. Let me denote that. Let me consider that (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore, 17² = r² + s² + t² + 2*104 => 289 = r² + s² + t² + 208 => r² + s² + t² = 289 - 208 = 81. So the sum of squares is 81. So we need three positive integers r, s, t with sum 17 and sum of squares 81. Let me check if such numbers exist. Let me think. Let me try possible triples. Let me think of numbers where squares add up to 81. For example, 9² = 81, so if one of them is 9, then the other two would have squares adding to 0, which is impossible. If two numbers are 9, then the third would be 81 - 2*81 = negative. Not possible. So maybe numbers less than 9. Let me think. Let me try 6² + 6² + 3² = 36 + 36 + 9 = 81. So 6, 6, 3. Let me check sum: 6 + 6 + 3 = 15 ≠ 17. Not good. How about 5² + 5² + 5² = 75. Not 81. 5² + 6² + 4² = 25 + 36 + 16 = 77. 5² + 6² + 5² = 25 + 36 + 25 = 86. 5² + 7² + 3² = 25 + 49 + 9 = 83. 5² + 8² + 0² = invalid. 4² + 5² + 6² = same as above. 4² + 4² + 7² = 16 + 16 + 49 = 81. So 4, 4, 7. Let me check sum: 4 + 4 + 7 = 15 ≠ 17. Not good. 3² + 6² + 6² = 9 + 36 + 36 = 81. Sum is 3 + 6 + 6 = 15. Still not 17.  How about 2² + 5² + 6² = 4 + 25 + 36 = 65. 2² + 6² + 7² = 4 + 36 + 49 = 89. 3² + 4² + 8² = 9 + 16 + 64 = 89. 3² + 5² + 7² = 9 + 25 + 49 = 83. 4² + 5² + 6² = 77.  So seems like there is no combination of three positive integers with sum 17 and sum of squares 81. Therefore, a=17 is invalid.

So maybe a=15 is the candidate. Let me check another a. Let me check a=17 didn't work. How about a=19? Let me check. rs + rt + st = (361 - 81)/2 = 280/2 = 140. And r + s + t = 19. Let me try to find triples. Let me think. Let me try to find numbers. Let me think of possible triples. Let me try 5, 6, 8: sum 19. Sum of products: 5*6 + 5*8 + 6*8 = 30 + 40 + 48 = 118. Not 140.  Let me think of equations. Let me use the same approach as before. (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore, 19² = r² + s² + t² + 2*140 => 361 = r² + s² + t² + 280 => r² + s² + t² = 81. Wait, same as before? Wait, 361 - 280 = 81. So sum of squares is 81. Wait, same as before. So same problem. So again, trying to find three numbers with sum 19 and sum of squares 81. Which seems impossible, since the maximum possible sum of squares for numbers adding to 19 would be if one number is 19 and others are 0, but they have to be positive integers. So for example, 1, 1, 17: sum of squares 1 + 1 + 289 = 291. Which is way larger than 81. So there's a contradiction here. Wait, that suggests that there is no solution for a=19. Therefore, maybe my previous approach is wrong? Wait, how come for a=15, we had sum of squares 81? Wait, for a=15, (r + s + t)^2 = 225, and rs + rt + st = 72. Therefore, r² + s² + t² = 225 - 2*72 = 225 - 144 = 81. So same as for a=19. But for a=15, we had triples (4,4,7) and (3,6,6) which sum to 15 and have sum of squares 16 + 16 + 49 = 81 and 9 + 36 + 36 = 81. So that works. But for a=19, we need sum of squares 81 with sum 19. Which is impossible because the minimal sum of squares for numbers adding to 19 is when numbers are as equal as possible. For example, 6, 6, 7: sum 19. Sum of squares: 36 + 36 + 49 = 121. Which is already larger than 81. Therefore, there is no solution for a=19. Therefore, my previous calculation for a=19 is invalid. Therefore, maybe there is a mistake here. Wait, but how come for a=15, the sum of squares is 81? Because for a=15, the sum of squares is 81, which is possible. For a=19, the sum of squares is 81, which is impossible. Therefore, there must be some mistake in my calculation. Wait, let me check again. For a=19, rs + rt + st = (a² - 81)/2 = (361 - 81)/2 = 280/2 = 140. Then (r + s + t)^2 = 19² = 361. Therefore, r² + s² + t² = 361 - 2*140 = 361 - 280 = 81. So that's correct. But how can three positive integers with sum 19 have sum of squares 81? It's impossible because even the minimal sum of squares for numbers adding to 19 is higher than 81. Therefore, there is no solution for a=19. Therefore, my previous calculation for a=15 is correct. Therefore, maybe a=15 is the only a where there are multiple c's? Wait, let me check another a. Let me check a=17 again. Wait, for a=17, rs + rt + st = (289 - 81)/2 = 208/2 = 104. Then (r + s + t)^2 = 17² = 289. Therefore, r² + s² + t² = 289 - 2*104 = 289 - 208 = 81. Same as before. Therefore, same problem. Therefore, for a=17, same as a=19, impossible. Therefore, only for a=15, we have possible triples. Wait, but let me check a=9. Wait, a must be greater than 9, since a² > 81. So a ≥ 10. But a must be odd, so starting from 11. But for a=11, we saw that there was no solution. For a=13, there was one solution. For a=15, two solutions. For a=17, no solution. For a=19, no solution. Let me check a=21. rs + rt + st = (441 - 81)/2 = 360/2 = 180. Then (r + s + t)^2 = 21² = 441. Therefore, r² + s² + t² = 441 - 2*180 = 441 - 360 = 81. Again same sum of squares. Which is impossible. Therefore, seems like only for a=15, we have possible triples. Therefore, maybe a=15 is the value that Steve told Jon, and there are two possible c's: 224 and 216. Therefore, the sum is 224 + 216 = 440. But let me check if there are other a's where there are multiple triples. Let me check a=17 again. Wait, no. How about a=13? For a=13, we had only one triple: 1,4,8. So c=64. So only one c. Therefore, Jon couldn't determine a and c initially because there are multiple a's with multiple c's? Wait, but initially, Jon couldn't determine a and c. So there must be multiple possible a's with multiple c's. Wait, but according to our previous analysis, only a=15 has multiple c's. Wait, but maybe there are other a's. Let me check a=12? Wait, a must be odd. So a=12 is even. Not allowed. a=14 is even. a=16 even. a=18 even. So only odd a's. So next after a=15 is a=17, which we saw no solution. How about a=7? Wait, a must be greater than 9. So no. How about a=11? Let me check again. For a=11, rs + rt + st = (121 - 81)/2 = 20. And r + s + t = 11. Let me try again. Maybe I missed some triple. Let me think. Let me try 2, 3, 6: sum 11. Sum of products: 2*3 + 2*6 + 3*6 = 6 + 12 + 18 = 36. Too big. 2, 2, 7: sum 11. Sum of products: 2*2 + 2*7 + 2*7 = 4 + 14 + 14 = 32. 1, 2, 8: sum 11. Sum of products: 1*2 + 1*8 + 2*8 = 2 + 8 + 16 = 26. 1, 3, 7: 1*3 + 1*7 + 3*7 = 3 + 7 + 21 = 31. 1, 4, 6: 1*4 + 1*6 + 4*6 = 4 + 6 + 24 = 34. 1, 5, 5: 1*5 + 1*5 + 5*5 = 5 + 5 + 25 = 35. 2, 4, 5: 2*4 + 2*5 + 4*5 = 8 + 10 + 20 = 38. 3, 3, 5: 3*3 + 3*5 + 3*5 = 9 + 15 + 15 = 39. 3, 4, 4: 3*4 + 3*4 + 4*4 = 12 + 12 + 16 = 40. So none of these give 20. Wait, but maybe there are other triples? For example, with 1, 1, 9: sum 11. Sum of products: 1*1 + 1*9 + 1*9 = 1 + 9 + 9 = 19. Close. If we have 1, 1, 9: sum of products 19. If we have 1, 2, 8: 26. So there is no way to get 20. Therefore, a=11 is invalid. Therefore, only a=13 and a=15. For a=13, only one triple. For a=15, two triples. Therefore, initially, Jon couldn't determine a and c because there are multiple possibilities. But when Steve told him a, there were still two possible c's. Therefore, Steve must have told Jon a=15, because for a=15, there are two c's. Therefore, the answer is 224 + 216 = 440. But let me check if there are other a's. Let me check a=21 again. Wait, no. How about a= something else? Let me check a= 23? rs + rt + st = (529 - 81)/2 = 448/2 = 224. Then (r + s + t)^2 = 23² = 529. Therefore, r² + s² + t² = 529 - 2*224 = 529 - 448 = 81. Again same problem. So impossible. Therefore, seems like only a=15 is possible. Therefore, the answer is 224 + 216 = 440. Wait, but let me check if there are other triples for a=15. For example, maybe (2, 5, 8)? Let me check. Sum is 2 + 5 + 8 = 15. Sum of products: 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. Not 72. So no. How about (2, 6, 7): sum 15. Sum of products: 2*6 + 2*7 + 6*7 = 12 + 14 + 42 = 68. Not 72. (3, 4, 8): 3 + 4 + 8 = 15. Sum of products: 3*4 + 3*8 + 4*8 = 12 + 24 + 32 = 68. (3, 5, 7): 3 + 5 + 7 = 15. Sum of products: 15 + 21 + 35 = 71. (4, 4, 7): 4 + 4 + 7 = 15. Sum of products: 16 + 28 + 28 = 72. (3, 6, 6): 3 + 6 + 6 = 15. Sum of products: 18 + 18 + 36 = 72. So only these two triples. Therefore, c is 2*4*4*7 = 224 and 2*3*6*6 = 216. Therefore, sum is 224 + 216 = 440. Therefore, the answer is 440. But let me check if there are other a's. For example, a= 9? No, since a must be greater than 9. How about a= 10? Wait, a must be odd. So no. Therefore, seems like the answer is 440. However, let me check if there are other a's where multiple triples exist. Let me check a= 17 again. Wait, no. How about a= 25? rs + rt + st = (625 - 81)/2 = 544/2 = 272. Then (r + s + t)^2 = 25² = 625. Therefore, r² + s² + t² = 625 - 2*272 = 625 - 544 = 81. Again same problem. So impossible. Therefore, only a=15. Therefore, the answer is 440. However, let me check if there are other triples for a=15. For example, (2, 2, 11): sum 15. Sum of products: 2*2 + 2*11 + 2*11 = 4 + 22 + 22 = 48. Not 72. (1, 2, 12): sum 15. Sum of products: 1*2 + 1*12 + 2*12 = 2 + 12 + 24 = 38. No. (1, 3, 11): 1*3 + 1*11 + 3*11 = 3 + 11 + 33 = 47. No. (1, 4, 10): 1*4 + 1*10 + 4*10 = 4 + 10 + 40 = 54. No. (1, 5, 9): 1*5 + 1*9 + 5*9 = 5 + 9 + 45 = 59. No. (1, 6, 8): 1*6 + 1*8 + 6*8 = 6 + 8 + 48 = 62. No. (2, 3, 10): 2*3 + 2*10 + 3*10 = 6 + 20 + 30 = 56. No. (2, 4, 9): 2*4 + 2*9 + 4*9 = 8 + 18 + 36 = 62. No. (2, 5, 8): 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. No. (2, 6, 7): 2*6 + 2*7 + 6*7 = 12 + 14 + 42 = 68. No. (3, 3, 9): 3*3 + 3*9 + 3*9 = 9 + 27 + 27 = 63. No. (3, 4, 8): 3*4 + 3*8 + 4*8 = 12 + 24 + 32 = 68. No. (3, 5, 7): 3*5 + 3*7 + 5*7 = 15 + 21 + 35 = 71. No. (4, 4, 7): 72. (3, 6, 6): 72. So only those two. Therefore, yes, only two c's. Therefore, the answer is 224 + 216 = 440. Therefore, the sum of the two possible values of c is 440.

But let me check once more if there are other a's. For example, a= 27? rs + rt + st = (729 - 81)/2 = 648/2 = 324. Then (r + s + t)^2 = 27² = 729. Therefore, r² + s² + t² = 729 - 2*324 = 729 - 648 = 81. Again same problem. So impossible. Therefore, no. Therefore, only a=15. Therefore, answer is 440.

But wait, the problem says that Jon initially couldn't determine a and c. So there must be multiple possible a's with multiple c's. But according to our analysis, only a=15 has multiple c's. However, maybe there are other a's where there are multiple triples. Let me check a= 15 and a= something else. Wait, let me check a= 9. Wait, a must be greater than 9. How about a= 13? For a=13, only one triple. So maybe there are multiple a's where there are multiple c's? Wait, but if there are multiple a's, then when Steve told Jon the value of a, Jon could have multiple c's. But according to the problem, Jon initially couldn't determine a and c, meaning that there are multiple possible a's and c's. But after Steve told him a, there were still two possible c's. Therefore, the value of a must be such that there are multiple c's. But if only a=15 has multiple c's, then that would mean that initially, there were multiple a's with multiple c's? Wait, but if only a=15 has multiple c's, then initially, Jon couldn't determine a and c because there is only one a (a=15) with multiple c's, but maybe there are other a's with unique c's? Wait, no. The problem says that Jon initially couldn't determine a and c. So there must be multiple possible polynomials, i.e., multiple (a, c) pairs. But if only a=15 has multiple c's, then there are multiple (a, c) pairs: for example, for a=13, there is one c=64; for a=15, there are two c's: 216 and 224. So total of three possible (a, c) pairs. Therefore, Jon couldn't determine a and c initially. Then, when Steve told him a, if a was 13, then c is uniquely determined. If a was 15, then c is not uniquely determined. Therefore, if Steve told Jon that a=15, then Jon would have two c's. But if Steve told Jon that a=13, then Jon would know c. Therefore, the problem says that after Steve told Jon the value of a, there were still two possible c's. Therefore, Steve must have told Jon a=15. Therefore, the answer is 216 + 224 = 440. Therefore, the answer is 440.

But let me check if there are other a's with multiple c's. For example, let me check a= 21. Wait, no. How about a= 7? No. How about a= 5? No. How about a= 17? No. How about a= 19? No. Therefore, only a=15. Therefore, the answer is 440.

But let me check once again for a= 15. The two triples are (3,6,6) and (4,4,7). Therefore, c=2*3*6*6= 2*108=216 and c=2*4*4*7=2*112=224. Therefore, sum is 216 + 224 = 440. Therefore, the answer is 440.

But wait, the problem says "the sum of the two possible values of c". Therefore, the answer is 440. However, I need to check if there are other possible triples for a=15. For example, maybe (2, 5, 8)? Wait, sum is 15. Sum of products is 2*5 + 2*8 + 5*8 = 10 + 16 + 40 = 66. Not 72. So no. How about (2, 3, 10)? Sum of products is 6 + 20 + 30 = 56. No. (1, 5, 9): 5 + 9 + 45 = 59. No. So only two triples. Therefore, answer is 440.

But let me check if there are other a's. For example, let me check a= 25. Wait, no. How about a= 15. Yes. Therefore, I think the answer is 440. However, I recall that in some similar problems, sometimes there are other possibilities. Let me check again.

Wait, let me think of another approach. Since we have:

From Vieta's formula:

r + s + t = a

rs + rt + st = (a² - 81)/2

rst = c/2

Let me denote that. Let me think of the equations. Let me consider that for given a, we can find possible triples. But since we have two equations, maybe we can find relations between them. Let me think. Let me consider that for three variables r, s, t, we can think of them as roots of the equation x³ - a x² + [(a² - 81)/2]x - (c/2) = 0. But since they are integers, maybe we can find relations. Alternatively, since we have r + s + t = a, and rs + rt + st = (a² - 81)/2. Let me denote that:

Let me compute (r + s + t)^2 = a² = r² + s² + t² + 2(rs + rt + st). Therefore:

a² = r² + s² + t² + 2*( (a² - 81)/2 )

Simplify:

a² = r² + s² + t² + a² - 81

Therefore:

0 = r² + s² + t² - 81 => r² + s² + t² = 81.

Wait, this is interesting! So regardless of a, we have that r² + s² + t² = 81. Wait, but earlier for a=15, we had that. For a=13, we had r² + s² + t² = 81? Let me check. For a=13, rs + rt + st = (169 - 81)/2 = 44. Then (r + s + t)^2 = 13² = 169. Therefore, r² + s² + t² = 169 - 2*44 = 169 - 88 = 81. Yes! So for any a, we have that r² + s² + t² = 81. Therefore, regardless of a, the sum of squares of the roots is 81. Therefore, the problem reduces to finding all triples of positive integers (r, s, t) with r + s + t = a, rs + rt + st = (a² - 81)/2, and r² + s² + t² = 81. Therefore, the key is that the sum of squares is 81. Therefore, we need to find all triples of positive integers (r, s, t) with r² + s² + t² = 81. Then, for each such triple, compute a = r + s + t, and then check if rs + rt + st = (a² - 81)/2. Wait, but since we already have that r² + s² + t² = 81, and a = r + s + t, then rs + rt + st = (a² - 81)/2. Therefore, for each triple (r, s, t) with r² + s² + t² = 81, we can compute a = r + s + t, and then c = 2rst. Therefore, the problem reduces to finding all triples of positive integers (r, s, t) with r² + s² + t² = 81, then for each such triple, compute a and c, and then find those a's which have multiple c's.

Therefore, let me find all triples (r, s, t) of positive integers with r² + s² + t² = 81. Let me list them.

First, note that since r, s, t are positive integers, each of them is at least 1. Also, since 81 is not too big, the maximum possible value for any of them is 9, since 10² = 100 > 81. So possible values for each variable are from 1 to 9.

Let me list all possible triples. Let me consider r ≤ s ≤ t to avoid duplicates.

Start with r=1:

Then s² + t² = 81 - 1 = 80. So find s and t with s ≤ t, s ≥ 1, t ≥ s, s² + t² = 80.

Possible s:

s=1: t² = 80 - 1 = 79. Not square.

s=2: t² = 80 - 4 = 76. Not square.

s=3: 80 - 9 = 71. No.

s=4: 80 - 16 = 64. Yes! t²=64 => t=8. So (1,4,8).

s=5: 80 - 25 = 55. No.

s=6: 80 - 36 = 44. No.

s=7: 80 - 49 = 31. No.

s=8: 80 - 64 = 16. t=4, but t ≥ s=8? No, since t=4 < s=8. So invalid. Therefore, only (1,4,8).

Next, r=2:

s² + t² = 81 - 4 = 77. Find s ≥ 2, t ≥ s.

s=2: t² = 77 - 4 = 73. No.

s=3: 77 - 9 = 68. No.

s=4: 77 - 16 = 61. No.

s=5: 77 - 25 = 52. No.

s=6: 77 - 36 = 41. No.

s=7: 77 - 49 = 28. No.

s=8: 77 - 64 = 13. No.

No solutions here.

r=3:

s² + t² = 81 - 9 = 72. s ≥ 3, t ≥ s.

s=3: t² = 72 - 9 = 63. No.

s=4: 72 - 16 = 56. No.

s=5: 72 - 25 = 47. No.

s=6: 72 - 36 = 36. t=6. So (3,6,6).

s=7: 72 - 49 = 23. No.

s=8: 72 - 64 = 8. No.

s=9: 72 - 81 = negative. No.

So only (3,6,6).

r=4:

s² + t² = 81 - 16 = 65. s ≥ 4, t ≥ s.

s=4: t² = 65 - 16 = 49. t=7. So (4,4,7).

s=5: 65 - 25 = 40. No.

s=6: 65 - 36 = 29. No.

s=7: 65 - 49 = 16. t=4 < s=7. Invalid.

s=8: 65 - 64 = 1. No.

So only (4,4,7).

r=5:

s² + t² = 81 - 25 = 56. s ≥ 5, t ≥ s.

s=5: t² = 56 - 25 = 31. No.

s=6: 56 - 36 = 20. No.

s=7: 56 - 49 = 7. No.

s=8: 56 - 64 = negative. No.

No solutions.

r=6:

s² + t² = 81 - 36 = 45. s ≥ 6, t ≥ s.

s=6: t² = 45 - 36 = 9. t=3 < s=6. Invalid.

s=7: 45 - 49 = negative. No.

No solutions.

r=7:

s² + t² = 81 - 49 = 32. s ≥ 7, t ≥ s.

s=7: t² = 32 - 49 = negative. No.

No solutions.

Similarly for r=8,9: even worse.

Therefore, the only triples are (1,4,8), (3,6,6), (4,4,7). Therefore, three triples.

Now, for each triple, compute a = r + s + t and c = 2rst.

For (1,4,8): a = 1 + 4 + 8 = 13, c = 2*1*4*8 = 64.

For (3,6,6): a = 3 + 6 + 6 = 15, c = 2*3*6*6 = 216.

For (4,4,7): a = 4 + 4 + 7 = 15, c = 2*4*4*7 = 224.

Therefore, there are three possible (a, c) pairs: (13, 64), (15, 216), (15, 224).

Therefore, initially, Jon couldn't determine a and c because there are multiple possibilities. However, when Steve told him the value of a, if a was 13, then c is uniquely 64. If a was 15, then there are two possible c's: 216 and 224. Therefore, the problem says that after Steve told him a, there were still two possible c's. Therefore, a must be 15, and the two possible c's are 216 and 224. Therefore, the sum is 216 + 224 = 440.

Therefore, the answer is 440.

**Final Answer**
\boxed{440}

To solve the problem, we are given a cubic polynomial:

$$
P(x) = 2x^3 - 2a x^2 + (a^2 - 81)x - c
$$

with all **positive integer roots** $ r, s, t $, and the coefficients $ a $ and $ c $ are also positive integers. Our goal is to find the **sum of the two possible values of $ c $** that remain after Steve reveals the value of $ a $, and Jon is still unable to determine a unique value of $ c $.

---

### Step 1: Use Vieta's Formulas

Given the polynomial:

$$
P(x) = 2x^3 - 2a x^2 + (a^2 - 81)x - c
$$

We can write it in factored form as:

$$
P(x) = 2(x - r)(x - s)(x - t)
$$

Expanding this, we get:

$$
P(x) = 2x^3 - 2(r + s + t)x^2 + 2(rs + rt + st)x - 2rst
$$

Comparing with the original polynomial:

- $ r + s + t = a $
- $ rs + rt + st = \frac{a^2 - 81}{2} $
- $ rst = \frac{c}{2} \Rightarrow c = 2rst $

---

### Step 2: Use the Identity for Sum of Squares

From the identity:

$$
(r + s + t)^2 = r^2 + s^2 + t^2 + 2(rs + rt + st)
$$

Substitute the known values:

$$
a^2 = r^2 + s^2 + t^2 + 2 \cdot \frac{a^2 - 81}{2}
$$

Simplify:

$$
a^2 = r^2 + s^2 + t^2 + a^2 - 81
\Rightarrow r^2 + s^2 + t^2 = 81
$$

So, the **sum of the squares of the roots is always 81**, regardless of the value of $ a $.

---

### Step 3: Find All Positive Integer Triples with $ r^2 + s^2 + t^2 = 81 $

We now find all **positive integer triples** $ (r, s, t) $ such that:

$$
r^2 + s^2 + t^2 = 81
$$

We consider all combinations with $ r \leq s \leq t $ to avoid duplicates.

- $ (1, 4, 8) $: $ 1^2 + 4^2 + 8^2 = 1 + 16 + 64 = 81 $
- $ (3, 6, 6) $: $ 9 + 36 + 36 = 81 $
- $ (4, 4, 7) $: $ 16 + 16 + 49 = 81 $

These are the only such triples.

---

### Step 4: Compute Corresponding $ a $ and $ c $

For each triple, compute:

- $ a = r + s + t $
- $ c = 2rst $

| Triple | $ a = r + s + t $ | $ c = 2rst $ |
|--------|---------------------|----------------|
| (1, 4, 8) | 13 | 64 |
| (3, 6, 6) | 15 | 216 |
| (4, 4, 7) | 15 | 224 |

So, the possible (a, c) pairs are:

- (13, 64)
- (15, 216)
- (15, 224)

---

### Step 5: Analyze the Problem Statement

- Initially, Jon couldn't determine $ a $ and $ c $, meaning there are **multiple possible (a, c) pairs**.
- After Steve reveals the value of $ a $, Jon still has **two possible values of $ c $**.

This implies that **Steve must have told Jon $ a = 15 $**, since that is the only value of $ a $ that corresponds to **two different values of $ c $**.

Thus, the two possible values of $ c $ are:

$$
c = 216 \quad \text{and} \quad c = 224
$$

---

### Final Step: Compute the Sum

$$
216 + 224 = \boxed{440}
$$

---

### ✅ Final Answer

$$
\boxed{440}
$$

Solution:
Let $r, s, t$ be the three positive integer roots of $P(x).$ Then by Vieta's formulas, \[\begin{aligned} r+s+t &= a, \\ rs+st+rt &= \frac{a^2-81}{2}, \\ rst &= \frac{c}{2}. \end{aligned}\]Substituting the first equation into the second to eliminate $a,$ we have \[rs+st+rt = \frac{(r+s+t)^2 - 81}{2} = \frac{(r^2+s^2+t^2) + 2(rs+st+rt) - 81}{2}.\]This simplifies to \[r^2 + s^2 + t^2 = 81.\]Therefore, each of $r, s, t$ lies in the set $\{1, 2, \ldots, 9\}.$ Assuming without loss of generality that $r \le s \le t,$ we have $81=r^2+s^2+t^2 \le 3t^2,$ so $t^2 \ge 27,$ and $t \ge 6.$ We take cases:

If $t = 6,$ then $r^2+s^2 = 81 - 6^2 = 45;$ the only solution where $r \le s \le 6$ is $(r, s) = (3, 6).$
If $t = 7,$ then $r^2+s^2 = 81-7^2 = 32;$ the only solution where $r \le s \le 7$ is $(r, s) = (4, 4).$
If $t = 8,$ then $r^2+s^2 = 81-8^2 = 17;$ the only solution where $r \le s \le 8$ is $(r, s) = (1, 4).$

Therefore, the possible sets of roots of such a polynomial are $(3, 6, 6), (4, 4, 7),$ and $(1, 4, 8).$ Calculating $a = r+s+t$ and $c=2rst$ for each set, we have $(a, c) = (15, 216), (15, 224), (13, 64).$

Since, given the value of $a,$ there are still two possible values of $c,$ it must be that $a = 15,$ since two of the pairs $(a, c)$ have $a = 15,$ but only one has $a = 13.$ Then the sum of the two possible values of $c$ is \[216 + 224 = \boxed{440}.\]


Explanation:
Okay, so I need to figure out how many integer coordinate points are strictly inside this rectangle with vertices at (5,4), (-5,4), (-5,-4), and (5,-4). Let me start by visualizing the rectangle. The rectangle spans from x = -5 to x = 5 and y = -4 to y = 4. But since the question is about points strictly inside, I need to exclude the boundaries.

First, let me recall that for a rectangle with sides parallel to the axes, the number of interior points can be calculated by considering the number of integer points along the x-axis and y-axis, then multiplying them. But since we need strictly interior points, we have to adjust for the boundaries.

Let me think. If the rectangle goes from x = -5 to x = 5, then the length along the x-axis is from -5 to 5. But since we want strictly interior points, we need to exclude the edges. So, for the x-coordinates, the interior points would be from -4 to 4, right? Because if you have a rectangle from -5 to 5, the interior would be points where x is greater than -5 and less than 5. But since we are dealing with integer coordinates, the integer points strictly inside would have x-values from -4 up to 4. Similarly for the y-coordinates, the original rectangle goes from -4 to 4, so the interior points would have y-values from -3 up to 3.

Wait, let me check that. If the rectangle includes the boundary points, then the total number of integer points on the boundary would be... but actually, the question is about strictly interior points. So, for the x-coordinate, the original rectangle includes x = -5 and x = 5. So, to be strictly inside, x must be greater than -5 and less than 5. Since x is integer, that means x can be from -4, -3, -2, -1, 0, 1, 2, 3, 4. Similarly for y-coordinate, original rectangle includes y = -4 and y = 4, so strictly inside would be y from -3, -2, -1, 0, 1, 2, 3.

Therefore, the number of interior points would be the number of x-values times the number of y-values. Let me calculate that.

For x-values: from -4 to 4 inclusive. How many integers is that? Let me compute. From -4 to 4, that's 4 - (-4) + 1 = 9? Wait, no. Wait, the formula for the number of integers from a to b inclusive is b - a + 1. So here, if a is -4 and b is 4, then 4 - (-4) + 1 = 4 + 4 + 1 = 9. So 9 x-values. Similarly for y-values: from -3 to 3 inclusive. So that's 3 - (-3) + 1 = 3 + 3 + 1 = 7. Therefore, total interior points would be 9 * 7 = 63.

Wait, but let me verify this with another approach. Maybe I can think of the total number of integer points in the rectangle including the boundary, then subtract the boundary points. But maybe that's more complicated? Let me try.

Total number of integer points in the rectangle including the boundary: For x from -5 to 5 inclusive, that's 5 - (-5) + 1 = 11 points. Similarly for y from -4 to 4 inclusive, that's 4 - (-4) + 1 = 9 points. So total points including boundary is 11 * 9 = 99.

Now, if I want the number of interior points, I need to subtract the boundary points. But how many boundary points are there? The boundary consists of the four sides. However, we have to be careful not to double count the corners.

Alternatively, maybe it's easier to use inclusion-exclusion. But maybe my first approach was correct? Let me check.

Alternatively, if I think of the interior points as those where x is strictly between -5 and 5, and y strictly between -4 and 4. So for x, integers from -4 to 4 inclusive, which is 9 numbers. For y, integers from -3 to 3 inclusive, which is 7 numbers. So 9*7=63.

But let me check with the other method. Total points including boundary: 11*9=99. Now, how many points are on the boundary? The boundary consists of the four sides. Let me compute the number of points on each side and then subtract overlaps.

First, the top side: y=4, from x=-5 to x=5. That's 11 points (from -5 to 5 inclusive). Similarly, the bottom side: y=-4, same thing, 11 points. The left side: x=-5, from y=-4 to y=4, which is 9 points. The right side: x=5, same as left, 9 points.

But now, if I add them up: top + bottom + left + right = 11 + 11 + 9 + 9 = 3 + 2*11 + 2*9? Wait, 11+11 is 22, 9+9 is 18, total 22+18=40. But this counts the corners twice. Each corner is shared by two sides. There are four corners. So total boundary points would be 40 - 4 = 36? Wait, because each corner is counted twice, once in each adjacent side. So we need to subtract 4 duplicates.

Therefore, total boundary points: 40 - 4 = 36. Therefore, the number of interior points would be total points (99) minus boundary points (36) = 63. Which matches the first method.

So that seems to confirm that the answer is 63.

But let me check again with another approach. For example, for the interior points, the x-coordinate must satisfy -5 < x < 5. Since x is integer, x can be -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 values. Similarly, y must satisfy -4 < y < 4, so y can be -3, -2, -1, 0, 1, 2, 3. That's 7 values. So 9*7=63.

Therefore, all methods lead to 63.

But let me think again if there's a possibility of mistake here. For example, sometimes when dealing with coordinates, people might confuse inclusive vs exclusive. Let me check for x. The original rectangle has vertices at x=-5 and x=5. So the rectangle includes all points where x is between -5 and 5, inclusive. So the interior would be x strictly between -5 and 5, so x from -4 to 4 inclusive. Similarly for y.

Alternatively, if I think of the rectangle as defined by the inequalities: -5 ≤ x ≤ 5 and -4 ≤ y ≤ 4. Then the interior would be -5 < x < 5 and -4 < y < 4. So for integer coordinates, x can be from -4 to 4 inclusive (since -5 < x < 5 implies x can be -4, -3,..., 4). Similarly for y.

So yes, that seems correct.

Alternatively, maybe I can think of the number of interior points as (length - 2) * (width - 2). Wait, but here length and width are in terms of units? Wait, if the rectangle is from -5 to 5 on x-axis, the length is 10 units. Similarly, on y-axis, from -4 to 4, length is 8 units. But if we think of the number of interior points, it's like (length - 2)*(width - 2). Wait, but that formula is for something else? Wait, no. Wait, if you have a grid, the number of interior points would be (number of points along x - 2) * (number of points along y - 2). Wait, but maybe not. Let me think.

Wait, if you have a rectangle that spans from x= a to x= b, inclusive, then the number of integer points along x is b - a + 1. If you want the interior points, you need to exclude the boundaries, so the number of interior points along x would be (b - a + 1) - 2 = b - a -1. Wait, for example, if a = -5 and b = 5, then number of interior points along x is 5 - (-5) -1 = 10 -1 = 9. Which matches our previous result. Similarly for y: from -4 to 4, number of interior points is 4 - (-4) -1 = 8 -1 = 7. So 9*7=63.

So that formula works. So in general, if you have a rectangle from x = a to x = b and y = c to y = d, then the number of interior points is (b - a -1)*(d - c -1).

But in this case, since the rectangle is from -5 to 5 on x, and -4 to 4 on y, then (5 - (-5) -1) * (4 - (-4) -1) = (10 -1)*(8 -1)= 9*7=63.

So that's another way to think about it.

Therefore, all methods lead to 63.

But just to be thorough, let me check with specific examples. Let's take a smaller rectangle where I can count manually.

Suppose we have a rectangle with vertices at (1,1), (-1,1), (-1,-1), (1,-1). So this is a square from x=-1 to 1 and y=-1 to 1. The interior points would be strictly inside, so x from -0 to 0? Wait, no. Wait, for x strictly between -1 and 1, so x=0. Similarly y=0. So only one point (0,0).

Using the formula: (1 - (-1) -1)*(1 - (-1) -1) = (2 -1)*(2 -1)=1*1=1. Correct.

Another example: rectangle from x=-2 to 2, y=-2 to 2. Then interior points would be x from -1,0,1 (3 points) and y from -1,0,1 (3 points). Total 9 points. Using formula: (2 - (-2) -1)*(2 - (-2) -1)= (4 -1)*(4 -1)= 3*3=9. Correct.

So formula works.

Therefore, applying it to original problem: (5 - (-5) -1)*(4 - (-4) -1)= (10 -1)*(8 -1)=9*7=63.

Therefore, answer is 63.

But just to make sure, let me think again. Suppose I list the x-values. From -4 to 4 inclusive. Let me count them: -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 numbers. Similarly for y: -3, -2, -1, 0, 1, 2, 3. That's 7 numbers. So 9*7=63.

Yes.

Alternatively, if I think of the grid. The rectangle is 10 units wide (from -5 to 5) and 8 units tall (from -4 to 4). But the number of interior points is (width - 2)*(height - 2). Wait, but here width is 10 units, so number of interior points along x is 10 - 2 = 8? Wait, no. Wait, if the width is 10 units, then the number of integer points along the width is 11 (from -5 to 5). Then the interior points would be 11 - 2 = 9. Similarly, height is 8 units, number of integer points is 9, so interior points would be 9 - 2 = 7. Therefore, 9*7=63.

Yes, that's another way to think about it. So the formula is (number of points along x - 2) * (number of points along y - 2).

Therefore, I think 63 is correct.

But just to be 100% sure, let me think of another way. Let me imagine the rectangle. The rectangle spans from x=-5 to x=5, so in terms of integer coordinates, the x-values go from -5 to 5 inclusive. Similarly for y.

If I want strictly interior points, then for x, it's all integers except -5 and 5. So total x-values: 11 total, minus 2 boundary points, so 9. Similarly for y: total y-values 9, minus 2 boundary points (y=-4 and y=4), so 7. Therefore, 9*7=63.

Yes, that's the same as before.

Therefore, I think the answer is 63.

**Final Answer**
The number of integer coordinate points strictly in the interior of the rectangular region is \boxed{63}.

To determine how many integer coordinate points lie **strictly inside** the rectangular region defined by the vertices $(5,4)$, $(-5,4)$, $(-5,-4)$, and $(5,-4)$, we proceed step by step.

---

### Step 1: Understand the Region

The rectangle is aligned with the coordinate axes and spans:

- **Horizontally** from $x = -5$ to $x = 5$
- **Vertically** from $y = -4$ to $y = 4$

This means the rectangle includes all points $(x, y)$ such that:
$$
-5 \leq x \leq 5 \quad \text{and} \quad -4 \leq y \leq 4
$$

However, the question asks for **strictly interior** points, i.e., points that are **not on the boundary** of the rectangle.

---

### Step 2: Determine the Range of Interior Coordinates

To be strictly inside the rectangle, the coordinates must satisfy:
$$
-5 < x < 5 \quad \text{and} \quad -4 < y < 4
$$

Since we are looking for **integer** coordinates, we consider the integer values within these ranges:

- For $x$: The integers strictly between $-5$ and $5$ are:
  $$
  -4, -3, -2, -1, 0, 1, 2, 3, 4 \quad \Rightarrow \quad 9 \text{ values}
  $$

- For $y$: The integers strictly between $-4$ and $4$ are:
  $$
  -3, -2, -1, 0, 1, 2, 3 \quad \Rightarrow \quad 7 \text{ values}
  $$

---

### Step 3: Compute the Total Number of Interior Points

Each interior point is a pair $(x, y)$ where $x$ is one of the 9 values and $y$ is one of the 7 values. Therefore, the total number of such points is:

$$
9 \times 7 = 63
$$

---

### Final Answer

$$
\boxed{63}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many integer coordinate points are strictly inside this rectangle with vertices at (5,4), (-5,4), (-5,-4), and (5,-4). Let me start by visualizing the rectangle. The rectangle spans from x = -5 to x = 5 and y = -4 to y = 4. But since the question is about points strictly inside, I need to exclude the boundaries.

First, let me recall that for a rectangle with sides parallel to the axes, the number of interior points can be calculated by considering the number of integer points along the x-axis and y-axis, then multiplying them. But since we need strictly interior points, we have to adjust for the boundaries.

Let me think. If the rectangle goes from x = -5 to x = 5, then the length along the x-axis is from -5 to 5. But since we want strictly interior points, we need to exclude the edges. So, for the x-coordinates, the interior points would be from -4 to 4, right? Because if you have a rectangle from -5 to 5, the interior would be points where x is greater than -5 and less than 5. But since we are dealing with integer coordinates, the integer points strictly inside would have x-values from -4 up to 4. Similarly for the y-coordinates, the original rectangle goes from -4 to 4, so the interior points would have y-values from -3 up to 3.

Wait, let me check that. If the rectangle includes the boundary points, then the total number of integer points on the boundary would be... but actually, the question is about strictly interior points. So, for the x-coordinate, the original rectangle includes x = -5 and x = 5. So, to be strictly inside, x must be greater than -5 and less than 5. Since x is integer, that means x can be from -4, -3, -2, -1, 0, 1, 2, 3, 4. Similarly for y-coordinate, original rectangle includes y = -4 and y = 4, so strictly inside would be y from -3, -2, -1, 0, 1, 2, 3.

Therefore, the number of interior points would be the number of x-values times the number of y-values. Let me calculate that.

For x-values: from -4 to 4 inclusive. How many integers is that? Let me compute. From -4 to 4, that's 4 - (-4) + 1 = 9? Wait, no. Wait, the formula for the number of integers from a to b inclusive is b - a + 1. So here, if a is -4 and b is 4, then 4 - (-4) + 1 = 4 + 4 + 1 = 9. So 9 x-values. Similarly for y-values: from -3 to 3 inclusive. So that's 3 - (-3) + 1 = 3 + 3 + 1 = 7. Therefore, total interior points would be 9 * 7 = 63.

Wait, but let me verify this with another approach. Maybe I can think of the total number of integer points in the rectangle including the boundary, then subtract the boundary points. But maybe that's more complicated? Let me try.

Total number of integer points in the rectangle including the boundary: For x from -5 to 5 inclusive, that's 5 - (-5) + 1 = 11 points. Similarly for y from -4 to 4 inclusive, that's 4 - (-4) + 1 = 9 points. So total points including boundary is 11 * 9 = 99.

Now, if I want the number of interior points, I need to subtract the boundary points. But how many boundary points are there? The boundary consists of the four sides. However, we have to be careful not to double count the corners.

Alternatively, maybe it's easier to use inclusion-exclusion. But maybe my first approach was correct? Let me check.

Alternatively, if I think of the interior points as those where x is strictly between -5 and 5, and y strictly between -4 and 4. So for x, integers from -4 to 4 inclusive, which is 9 numbers. For y, integers from -3 to 3 inclusive, which is 7 numbers. So 9*7=63.

But let me check with the other method. Total points including boundary: 11*9=99. Now, how many points are on the boundary? The boundary consists of the four sides. Let me compute the number of points on each side and then subtract overlaps.

First, the top side: y=4, from x=-5 to x=5. That's 11 points (from -5 to 5 inclusive). Similarly, the bottom side: y=-4, same thing, 11 points. The left side: x=-5, from y=-4 to y=4, which is 9 points. The right side: x=5, same as left, 9 points.

But now, if I add them up: top + bottom + left + right = 11 + 11 + 9 + 9 = 3 + 2*11 + 2*9? Wait, 11+11 is 22, 9+9 is 18, total 22+18=40. But this counts the corners twice. Each corner is shared by two sides. There are four corners. So total boundary points would be 40 - 4 = 36? Wait, because each corner is counted twice, once in each adjacent side. So we need to subtract 4 duplicates.

Therefore, total boundary points: 40 - 4 = 36. Therefore, the number of interior points would be total points (99) minus boundary points (36) = 63. Which matches the first method.

So that seems to confirm that the answer is 63.

But let me check again with another approach. For example, for the interior points, the x-coordinate must satisfy -5 < x < 5. Since x is integer, x can be -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 values. Similarly, y must satisfy -4 < y < 4, so y can be -3, -2, -1, 0, 1, 2, 3. That's 7 values. So 9*7=63.

Therefore, all methods lead to 63.

But let me think again if there's a possibility of mistake here. For example, sometimes when dealing with coordinates, people might confuse inclusive vs exclusive. Let me check for x. The original rectangle has vertices at x=-5 and x=5. So the rectangle includes all points where x is between -5 and 5, inclusive. So the interior would be x strictly between -5 and 5, so x from -4 to 4 inclusive. Similarly for y.

Alternatively, if I think of the rectangle as defined by the inequalities: -5 ≤ x ≤ 5 and -4 ≤ y ≤ 4. Then the interior would be -5 < x < 5 and -4 < y < 4. So for integer coordinates, x can be from -4 to 4 inclusive (since -5 < x < 5 implies x can be -4, -3,..., 4). Similarly for y.

So yes, that seems correct.

Alternatively, maybe I can think of the number of interior points as (length - 2) * (width - 2). Wait, but here length and width are in terms of units? Wait, if the rectangle is from -5 to 5 on x-axis, the length is 10 units. Similarly, on y-axis, from -4 to 4, length is 8 units. But if we think of the number of interior points, it's like (length - 2)*(width - 2). Wait, but that formula is for something else? Wait, no. Wait, if you have a grid, the number of interior points would be (number of points along x - 2) * (number of points along y - 2). Wait, but maybe not. Let me think.

Wait, if you have a rectangle that spans from x= a to x= b, inclusive, then the number of integer points along x is b - a + 1. If you want the interior points, you need to exclude the boundaries, so the number of interior points along x would be (b - a + 1) - 2 = b - a -1. Wait, for example, if a = -5 and b = 5, then number of interior points along x is 5 - (-5) -1 = 10 -1 = 9. Which matches our previous result. Similarly for y: from -4 to 4, number of interior points is 4 - (-4) -1 = 8 -1 = 7. So 9*7=63.

So that formula works. So in general, if you have a rectangle from x = a to x = b and y = c to y = d, then the number of interior points is (b - a -1)*(d - c -1).

But in this case, since the rectangle is from -5 to 5 on x, and -4 to 4 on y, then (5 - (-5) -1) * (4 - (-4) -1) = (10 -1)*(8 -1)= 9*7=63.

So that's another way to think about it.

Therefore, all methods lead to 63.

But just to be thorough, let me check with specific examples. Let's take a smaller rectangle where I can count manually.

Suppose we have a rectangle with vertices at (1,1), (-1,1), (-1,-1), (1,-1). So this is a square from x=-1 to 1 and y=-1 to 1. The interior points would be strictly inside, so x from -0 to 0? Wait, no. Wait, for x strictly between -1 and 1, so x=0. Similarly y=0. So only one point (0,0).

Using the formula: (1 - (-1) -1)*(1 - (-1) -1) = (2 -1)*(2 -1)=1*1=1. Correct.

Another example: rectangle from x=-2 to 2, y=-2 to 2. Then interior points would be x from -1,0,1 (3 points) and y from -1,0,1 (3 points). Total 9 points. Using formula: (2 - (-2) -1)*(2 - (-2) -1)= (4 -1)*(4 -1)= 3*3=9. Correct.

So formula works.

Therefore, applying it to original problem: (5 - (-5) -1)*(4 - (-4) -1)= (10 -1)*(8 -1)=9*7=63.

Therefore, answer is 63.

But just to make sure, let me think again. Suppose I list the x-values. From -4 to 4 inclusive. Let me count them: -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 numbers. Similarly for y: -3, -2, -1, 0, 1, 2, 3. That's 7 numbers. So 9*7=63.

Yes.

Alternatively, if I think of the grid. The rectangle is 10 units wide (from -5 to 5) and 8 units tall (from -4 to 4). But the number of interior points is (width - 2)*(height - 2). Wait, but here width is 10 units, so number of interior points along x is 10 - 2 = 8? Wait, no. Wait, if the width is 10 units, then the number of integer points along the width is 11 (from -5 to 5). Then the interior points would be 11 - 2 = 9. Similarly, height is 8 units, number of integer points is 9, so interior points would be 9 - 2 = 7. Therefore, 9*7=63.

Yes, that's another way to think about it. So the formula is (number of points along x - 2) * (number of points along y - 2).

Therefore, I think 63 is correct.

But just to be 100% sure, let me think of another way. Let me imagine the rectangle. The rectangle spans from x=-5 to x=5, so in terms of integer coordinates, the x-values go from -5 to 5 inclusive. Similarly for y.

If I want strictly interior points, then for x, it's all integers except -5 and 5. So total x-values: 11 total, minus 2 boundary points, so 9. Similarly for y: total y-values 9, minus 2 boundary points (y=-4 and y=4), so 7. Therefore, 9*7=63.

Yes, that's the same as before.

Therefore, I think the answer is 63.

**Final Answer**
The number of integer coordinate points strictly in the interior of the rectangular region is \boxed{63}.

To determine how many integer coordinate points lie **strictly inside** the rectangular region defined by the vertices $(5,4)$, $(-5,4)$, $(-5,-4)$, and $(5,-4)$, we proceed step by step.

---

### Step 1: Understand the Region

The rectangle is aligned with the coordinate axes and spans:

- **Horizontally** from $x = -5$ to $x = 5$
- **Vertically** from $y = -4$ to $y = 4$

This means the rectangle includes all points $(x, y)$ such that:
$$
-5 \leq x \leq 5 \quad \text{and} \quad -4 \leq y \leq 4
$$

However, the question asks for **strictly interior** points, i.e., points that are **not on the boundary** of the rectangle.

---

### Step 2: Determine the Range of Interior Coordinates

To be strictly inside the rectangle, the coordinates must satisfy:
$$
-5 < x < 5 \quad \text{and} \quad -4 < y < 4
$$

Since we are looking for **integer** coordinates, we consider the integer values within these ranges:

- For $x$: The integers strictly between $-5$ and $5$ are:
  $$
  -4, -3, -2, -1, 0, 1, 2, 3, 4 \quad \Rightarrow \quad 9 \text{ values}
  $$

- For $y$: The integers strictly between $-4$ and $4$ are:
  $$
  -3, -2, -1, 0, 1, 2, 3 \quad \Rightarrow \quad 7 \text{ values}
  $$

---

### Step 3: Compute the Total Number of Interior Points

Each interior point is a pair $(x, y)$ where $x$ is one of the 9 values and $y$ is one of the 7 values. Therefore, the total number of such points is:

$$
9 \times 7 = 63
$$

---

### Final Answer

$$
\boxed{63}
$$

Solution:
The rectangular region is 10 units by 8 units, resulting in an 8 by 6 rectangular region in the interior, which forms a 9-by-7 array of lattice points. That's $\boxed{63}$ points with integer coordinates, as shown in the figure. [asy]
import olympiad; size(150); defaultpen(linewidth(0.8));
add(grid(10,8));
draw((1,1)--(9,1)--(9,7)--(1,7)--cycle,linewidth(1.2));
[/asy] Note: We are counting points, not squares. It is a common mistake to count the interior squares, for 48, instead of interior lattice points, which gives the correct answer of 63.