Explanation:
Okay, so I need to find the smallest positive integer value of 'a' such that the sum of the geometric series 4 + 12/a + 36/a² + ... is a perfect square. Let me start by recalling the formula for the sum of an infinite geometric series.

The formula is S = a₁ / (1 - r), where a₁ is the first term and r is the common ratio, provided that |r| < 1.

First, let me identify the first term and the common ratio of the given series. The first term, a₁, is clearly 4. Now, the second term is 12/a. To find the common ratio r, I can divide the second term by the first term. So, r = (12/a) / 4 = (12/a) * (1/4) = 3/a.

Let me check if this ratio makes sense. The third term should be the second term multiplied by r. Let's see: second term is 12/a, multiplied by r = 3/a gives (12/a)*(3/a) = 36/a², which matches the third term given. So that's correct.

So the common ratio r is 3/a. Now, for the series to converge, we need |r| < 1. Since a is a positive integer, and the terms are positive, we can drop the absolute value and just require that r < 1. Therefore, 3/a < 1. Solving for a, we get a > 3. So a must be an integer greater than 3.

Now, the sum S of the series is given by S = a₁ / (1 - r) = 4 / (1 - 3/a). Let me simplify that denominator.

First, 1 - 3/a is equal to (a - 3)/a. Therefore, S = 4 / [(a - 3)/a] = 4 * [a / (a - 3)] = (4a)/(a - 3).

So the sum S is (4a)/(a - 3). The problem states that this sum must be a perfect square. Therefore, we need (4a)/(a - 3) to be a perfect square.

Our task is to find the smallest positive integer a > 3 such that (4a)/(a - 3) is a perfect square.

Let me denote S = (4a)/(a - 3) = k², where k is a positive integer.

So, we have the equation:

(4a)/(a - 3) = k².

Let me solve this equation for a in terms of k.

Multiply both sides by (a - 3):

4a = k²(a - 3)

Expanding the right side:

4a = k²a - 3k²

Bring all terms to one side:

4a - k²a + 3k² = 0

Factor out a:

a(4 - k²) + 3k² = 0

Then:

a(4 - k²) = -3k²

Multiply both sides by -1:

a(k² - 4) = 3k²

Therefore:

a = (3k²)/(k² - 4)

So, a must be equal to 3k² divided by (k² - 4). Since a must be a positive integer, we need (k² - 4) to divide 3k², and also k² - 4 must be positive because a is positive.

So, k² - 4 > 0 => k² > 4 => k > 2. Since k is a positive integer, k ≥ 3.

Therefore, k is an integer greater than or equal to 3.

Our goal is to find the smallest a, so we need to find the smallest k ≥ 3 such that (k² - 4) divides 3k², and then compute a from that.

Alternatively, since a must be an integer, (k² - 4) must divide 3k². Let me think about how to approach this.

Let me write a = 3k² / (k² - 4). Let me try to simplify this expression.

Let me perform the division:

a = 3k² / (k² - 4) = 3 + [12 / (k² - 4)]

Wait, let me check that. Let me do polynomial division or manipulate the expression:

Let me write 3k² as 3(k² - 4) + 12.

Yes:

3(k² - 4) = 3k² - 12. Therefore, 3k² = 3(k² - 4) + 12.

Therefore, substituting back into a:

a = [3(k² - 4) + 12]/(k² - 4) = 3 + 12/(k² - 4)

Therefore, a = 3 + 12/(k² - 4).

Since a must be an integer, 12/(k² - 4) must be an integer. Therefore, (k² - 4) must be a positive divisor of 12.

So, k² - 4 divides 12. Also, since k² - 4 is positive, as established earlier, k² - 4 is a positive divisor of 12.

Therefore, let me list all positive divisors of 12.

The positive divisors of 12 are: 1, 2, 3, 4, 6, 12.

Therefore, k² - 4 must be one of these numbers.

So, k² - 4 ∈ {1, 2, 3, 4, 6, 12}

Therefore, k² ∈ {5, 6, 7, 8, 10, 16}

But k is an integer ≥ 3, so let's check which of these values of k² are perfect squares.

Looking at the numbers:

5: not a perfect square.

6: not a perfect square.

7: not a perfect square.

8: not a perfect square.

10: not a perfect square.

16: yes, 4² = 16.

So, only k² = 16 is a perfect square in that list. Therefore, k² - 4 = 12, so k² = 16, so k = 4.

Wait, but that seems like only one possible value? But maybe I missed something here. Let me check again.

Wait, the divisors of 12 are 1, 2, 3, 4, 6, 12. So k² - 4 must be one of these. Therefore, k² must be 5, 6, 7, 8, 10, or 16.

But only 16 is a perfect square. Therefore, k = 4.

But then, if k = 4, then a = 3 + 12/(k² - 4) = 3 + 12/12 = 3 + 1 = 4.

But wait, earlier we had that a must be greater than 3. So a = 4 is allowed.

But let me check if this works. Let me compute the sum S when a = 4.

Sum S = 4a/(a - 3) = 4*4/(4 - 3) = 16/1 = 16. Which is 4², a perfect square.

So that works. But the question says "the smallest possible value of a where a is a positive integer". But a = 4 is possible. However, wait, the problem says "the geometric series 4 + 12/a + 36/a² + ...". If a = 4, then the series is 4 + 12/4 + 36/16 + ... which is 4 + 3 + 2.25 + ... and the sum is 16.

But the problem says "the smallest possible value of a". But maybe there are other values of k where k² - 4 divides 12, but k² is not necessarily in the list? Wait, no. Because k² - 4 must divide 12, so k² - 4 is a positive divisor of 12, so k² must be 4 + d, where d is a positive divisor of 12.

But maybe there are other possibilities? Wait, but if k² - 4 divides 12, then k² - 4 must be a positive integer divisor of 12. Therefore, the possible values of k² - 4 are 1, 2, 3, 4, 6, 12. Therefore, k² is 5, 6, 7, 8, 10, 16.

But only 16 is a perfect square. Therefore, only k = 4. Therefore, only a = 4.

But that seems too easy. However, the problem says "the smallest possible value of a where a is a positive integer". If a = 4 is possible, then that would be the answer. But maybe I made a mistake here? Let me check again.

Wait, let me check if there are other divisors. Wait, maybe negative divisors? But since k² - 4 is positive, we don't need to consider negative divisors.

Alternatively, perhaps my approach is missing something? Let me think again.

We had that a = 3k²/(k² - 4). So, for a to be integer, (k² - 4) must divide 3k².

But maybe there are other ways for (k² - 4) to divide 3k² without necessarily k² - 4 dividing 12? Wait, no. Because earlier, I rewrote a as 3 + 12/(k² - 4). Therefore, 12/(k² - 4) must be integer. Therefore, k² - 4 divides 12. Therefore, my previous reasoning holds.

Therefore, only k = 4 gives a valid solution. Therefore, a = 4.

But let me check if there are other possibilities. For example, maybe k² - 4 divides 3k², but not necessarily divides 12? Wait, but in the expression a = 3 + 12/(k² - 4), the term 12/(k² - 4) must be integer. Therefore, k² - 4 divides 12. Therefore, my previous conclusion is correct.

But then, if a = 4 is possible, then why does the problem ask for the smallest possible value? Maybe there is a mistake here? Let me check with a = 4.

Original series: 4 + 12/4 + 36/16 + ...

Which is 4 + 3 + 2.25 + 1.6875 + ...

Sum is 16, which is 4². So that's correct.

But maybe there is a smaller a? Wait, a must be greater than 3. The possible a's are integers starting from 4. So 4 is the smallest possible.

But the problem says "the smallest possible value of a where a is a positive integer". So 4 is the answer? But maybe I made a mistake in assuming that k² - 4 divides 12? Let me check again.

Wait, let me think again.

We had:

a = 3k²/(k² - 4).

So, for a to be integer, (k² - 4) must divide 3k².

But perhaps there are other values of k where (k² - 4) divides 3k², even if (k² - 4) does not divide 12?

Wait, but earlier, I rewrote a as 3 + 12/(k² - 4). Therefore, 12/(k² - 4) must be integer. Therefore, k² - 4 divides 12. Therefore, my previous reasoning is correct.

But let me check with k = 5.

If k = 5, then k² = 25. Then, k² - 4 = 21. Then, a = 3*(25)/21 = 75/21 = 25/7 ≈ 3.57, which is not integer.

Similarly, k = 3. Let me check k = 3.

k = 3, then k² = 9. Then, k² - 4 = 5. Then, a = 3*9 / 5 = 27/5 = 5.4, not integer.

k = 4, as before, gives a = 4.

k = 5, as above, gives non-integer.

k = 6: k² = 36, k² - 4 = 32. Then, a = 3*36 / 32 = 108/32 = 27/8, not integer.

k = 2: but k must be ≥ 3.

k = 1: no.

k = 0: no.

So, only k = 4 gives integer a. Therefore, a = 4.

But the problem says "the smallest possible value of a where a is a positive integer". So 4 is the answer?

But maybe there's a mistake here. Let me check again.

Wait, let me think if there are other possibilities where k² - 4 divides 3k², but not necessarily divides 12.

Wait, for example, suppose that k² - 4 divides 3k². Let me write that as:

Let d = k² - 4. Then, d divides 3k². Also, since d = k² - 4, then k² = d + 4. Therefore, substituting into 3k², we get 3(d + 4) = 3d + 12. Therefore, d divides 3d + 12. Which implies that d divides 12. Because if d divides 3d + 12, then d divides (3d + 12) - 3*d = 12.

Therefore, d divides 12. Therefore, my previous conclusion is correct. Therefore, d must be a positive divisor of 12. Therefore, k² - 4 is a positive divisor of 12. Therefore, only possible values of k² - 4 are 1, 2, 3, 4, 6, 12. Therefore, only k² = 5, 6, 7, 8, 10, 16.

Only k² = 16 is a perfect square, so k = 4. Therefore, a = 4.

Therefore, the answer is 4? But maybe the problem is expecting a different answer? Let me check again.

Wait, let me check if there are other divisors. For example, if d divides 12, but d could be negative? But since d = k² - 4, and k is integer ≥ 3, then d is positive. Therefore, no.

Alternatively, maybe I made a mistake in the formula for the sum? Let me check again.

Original series: 4 + 12/a + 36/a² + ...

First term a₁ = 4. Common ratio r = (12/a)/4 = 3/a.

Sum S = a₁ / (1 - r) = 4 / (1 - 3/a) = 4 / [(a - 3)/a] = 4a / (a - 3).

Yes, that's correct.

So S = 4a/(a - 3).

So if a = 4, S = 16/1 = 16, which is 4².

But maybe there is a smaller a? Wait, a must be greater than 3. So the next possible integer is 4. Therefore, 4 is the minimal.

But maybe the problem is in another interpretation? Let me check the problem again.

"Consider the geometric series 4 + 12/a + 36/a² + ... . If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?"

So, the problem says "the smallest possible value of a where a is a positive integer". So if a = 4 works, then that's the answer.

But maybe there is a mistake here? Let me check with a = 5.

If a = 5, then S = 4*5/(5 - 3) = 20/2 = 10. 10 is not a perfect square.

a = 6: 4*6/(6 - 3) = 24/3 = 8, not a perfect square.

a = 7: 4*7/(7 - 3) = 28/4 = 7, not a perfect square.

a = 8: 4*8/(8 - 3) = 32/5 = 6.4, not integer.

a = 9: 4*9/(9 - 3) = 36/6 = 6, not a perfect square.

a = 10: 4*10/(10 - 3) = 40/7 ≈ 5.71, not integer.

a = 12: 4*12/(12 - 3) = 48/9 = 16/3 ≈ 5.33, not integer.

a = 16: 4*16/(16 - 3) = 64/13 ≈ 4.92, not integer.

Wait, but according to our previous reasoning, only a = 4 works.

But let me check for k = something else. Wait, maybe there are other k where k² - 4 divides 12? For example, if k² - 4 is 1, 2, 3, 4, 6, 12.

But k² - 4 = 1 => k² = 5, which is not a square.

k² - 4 = 2 => k² = 6, not square.

k² - 4 = 3 => k² = 7, nope.

k² - 4 = 4 => k² = 8, nope.

k² - 4 = 6 => k² = 10, nope.

k² - 4 = 12 => k² = 16, which is 4².

Therefore, only k = 4.

Therefore, only a = 4.

But maybe there's another approach where a is not necessarily derived from k? Let me think.

Suppose that S = 4a/(a - 3) is a perfect square. Let me denote S = m², where m is a positive integer.

So, 4a/(a - 3) = m².

Let me rearrange this equation:

4a = m²(a - 3)

=> 4a = m²a - 3m²

=> 4a - m²a = -3m²

=> a(4 - m²) = -3m²

=> a = (3m²)/(m² - 4)

Which is the same as before.

So, same result. Therefore, same conclusion.

Therefore, the only possible value is m = 4, leading to a = 4.

But maybe m is not an integer? Wait, no, because S is a perfect square, so m must be integer.

Alternatively, maybe m is rational? But since S is a perfect square, and S is equal to 4a/(a - 3), which is rational because a is integer. Therefore, if S is a perfect square, then it must be a square of a rational number. However, since the problem says "a perfect square", it's likely referring to an integer perfect square.

But let me check. Suppose that S is a perfect square in the sense of rational numbers. For example, maybe S is (p/q)² where p and q are integers. However, the problem says "the sum is a perfect square", and in math competitions, usually, unless specified otherwise, perfect square refers to integer perfect squares.

Therefore, assuming that S must be an integer perfect square. Therefore, our previous reasoning holds.

Therefore, the answer is a = 4.

But let me check again with a = 4.

Sum is 16, which is 4². So that's correct.

But maybe the problem is expecting a different answer? Maybe I made a mistake in the initial steps?

Wait, let me check the problem statement again.

"Consider the geometric series 4 + 12/a + 36/a² + ... . If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?"

Wait, the series is 4 + 12/a + 36/a² + ... So the first term is 4, second term is 12/a, third term is 36/a², etc.

So the ratio between the second term and the first term is (12/a)/4 = 3/a. Then the ratio between the third term and the second term is (36/a²)/(12/a) = 3/a. So the common ratio is indeed 3/a.

Therefore, the sum is 4/(1 - 3/a) = 4a/(a - 3).

So that's correct.

Therefore, the answer is 4.

But maybe the problem is from a source where the answer is different? Wait, maybe I need to check if there are other values of a where 4a/(a - 3) is a perfect square.

Let me suppose that 4a/(a - 3) is a perfect square. Let me denote that as m².

So, 4a/(a - 3) = m².

Let me rearrange this equation:

Multiply both sides by (a - 3):

4a = m²(a - 3)

=> 4a = m²a - 3m²

=> 4a - m²a = -3m²

=> a(4 - m²) = -3m²

=> a = (3m²)/(m² - 4)

So, same as before.

Therefore, for a to be positive integer, m² - 4 must divide 3m² and m² - 4 must be positive.

So m² - 4 divides 3m².

Let me think of m² - 4 divides 3m².

Let me write m² - 4 divides 3m².

Which implies that m² - 4 divides 3m² - 3*(m² - 4) = 3m² - 3m² + 12 = 12.

Therefore, m² - 4 divides 12.

Therefore, m² - 4 is a positive divisor of 12.

Therefore, same conclusion as before.

Therefore, m² - 4 ∈ {1, 2, 3, 4, 6, 12}

Therefore, m² ∈ {5, 6, 7, 8, 10, 16}

Only m² = 16 is a perfect square. Therefore, m = 4.

Therefore, a = 3*(16)/(16 - 4) = 48/12 = 4.

Therefore, the answer is 4.

But maybe the problem is expecting a different answer? Let me check with m = something else.

Wait, suppose m² - 4 divides 12, but m² - 4 is negative? But since m² - 4 must be positive, as we saw earlier.

Alternatively, maybe m is not integer? But if m is rational, then S is a square of rational number. But the problem says "perfect square", which is ambiguous. However, in most cases, unless specified, perfect square refers to integer.

But let me check if there are other possibilities where S is a square of a rational number.

Suppose that S = (p/q)², where p and q are integers with no common factors. Then, 4a/(a - 3) = p²/q².

Cross multiplying: 4a q² = p²(a - 3).

But this seems more complicated. However, since a is integer, maybe there are solutions where S is a rational square but not integer. However, the problem says "the sum is a perfect square". If they mean a square of a rational number, then there could be more solutions.

But given that the problem is likely from a competition or textbook problem, and given that the answer is expected to be an integer, and the sum is 16 when a = 4, which is an integer perfect square, I think that the answer is 4.

But maybe there is a mistake here. Let me check with a = 5.

If a = 5, then S = 4*5/(5 - 3) = 20/2 = 10. 10 is not a perfect square.

a = 6: 24/3 = 8, not square.

a = 7: 28/4 = 7, not square.

a = 8: 32/5 = 6.4, not integer.

a = 9: 36/6 = 6, not square.

a = 10: 40/7 ≈ 5.71, not integer.

a = 12: 48/9 = 16/3 ≈ 5.33, not integer.

a = 16: 64/13 ≈ 4.92, not integer.

a = 20: 80/17 ≈ 4.7, not integer.

a = 24: 96/21 = 32/7 ≈ 4.57, not integer.

a = 3: but a must be greater than 3.

a = 4: 16, which is square.

So, seems like a = 4 is the only possible.

But maybe there is another way where m² - 4 divides 12, but m is not integer? For example, if m is rational.

But if m is rational, then m² is rational. However, since S = 4a/(a - 3) must be rational, which it is, but the problem says "perfect square". If they mean square of rational number, then there could be more solutions.

But let me think. Suppose that S is a square of a rational number. Then, 4a/(a - 3) = (p/q)².

Cross multiplying: 4a q² = p²(a - 3).

But this equation may have solutions for other a. However, since the problem says "the sum is a perfect square", and given that the answer is likely to be an integer, and given that a = 4 works, maybe that's the answer.

Alternatively, maybe the problem is from a source where the answer is different. Wait, let me check if there is a mistake in my calculation.

Wait, let me check for a = 5.

If a = 5, then the sum is 10. Is 10 a perfect square? No.

a = 6: 8, nope.

a = 7: 7, nope.

a = 8: 32/5, nope.

a = 9: 6, nope.

a = 10: 40/7, nope.

a = 12: 16/3, nope.

a = 15: 4*15/(15 - 3) = 60/12 = 5, not square.

a = 16: 64/13, nope.

a = 20: 80/17, nope.

a = 24: 96/21 = 32/7, nope.

a = 3: invalid.

a = 4: 16, which is square.

So, seems like only a = 4.

But maybe there is another value of a where 4a/(a - 3) is a perfect square. Let me think algebraically.

Let me set 4a/(a - 3) = m².

Let me write this as 4a = m²(a - 3).

Let me rearrange:

4a = m²a - 3m²

=> 4a - m²a = -3m²

=> a(4 - m²) = -3m²

=> a = (3m²)/(m² - 4)

So, for a to be integer, m² - 4 must divide 3m².

Let me think of m² - 4 divides 3m².

Let me write m² - 4 divides 3m².

Let me denote d = m² - 4. Then, d divides 3m².

But since d = m² - 4, then m² = d + 4.

Therefore, substituting into 3m², we get 3(d + 4) = 3d + 12.

Therefore, d divides 3d + 12 => d divides 12.

Therefore, d is a positive divisor of 12.

Therefore, same conclusion.

Therefore, only possible d is 1, 2, 3, 4, 6, 12.

Therefore, m² = d + 4.

So, m² = 5, 6, 7, 8, 10, 16.

Only m² = 16 is a perfect square.

Therefore, m = 4, leading to a = 3*(16)/(16 - 4) = 48/12 = 4.

Therefore, the answer is 4.

Therefore, I think the answer is 4.

But maybe the problem is in another interpretation? For example, maybe the series is finite? But the problem says "geometric series", which usually implies infinite unless specified otherwise.

Alternatively, maybe the problem is from a source where the answer is different. But given the calculations, I think 4 is correct.

Alternatively, maybe I made a mistake in the initial formula? Let me check again.

Sum of infinite geometric series is a₁/(1 - r).

First term a₁ = 4.

Common ratio r = 3/a.

Therefore, sum S = 4 / (1 - 3/a) = 4 / [(a - 3)/a] = 4a/(a - 3).

Yes, that's correct.

Therefore, I think the answer is 4.

But to be thorough, let me check if there are other values of a where 4a/(a - 3) is a perfect square.

Suppose that 4a/(a - 3) = k².

Let me rearrange this equation:

Let me write it as 4a = k²(a - 3)

=> 4a = k²a - 3k²

=> 4a - k²a = -3k²

=> a(4 - k²) = -3k²

=> a = 3k²/(k² - 4)

So, for a to be integer, k² - 4 must divide 3k².

Let me think of k as integer ≥ 3.

Let me try k = 5:

a = 3*(25)/(25 - 4) = 75/21 = 25/7 ≈ 3.57, not integer.

k = 6: 3*36/(36 - 4) = 108/32 = 27/8 ≈ 3.375, not integer.

k = 7: 3*49/(49 - 4) = 147/45 = 49/15 ≈ 3.266..., nope.

k = 8: 3*64/(64 - 4) = 192/60 = 16/5 = 3.2, nope.

k = 9: 3*81/(81 - 4) = 243/77 ≈ 3.155, nope.

k = 10: 3*100/(100 - 4) = 300/96 = 25/8 = 3.125, nope.

k = 12: 3*144/(144 - 4) = 432/140 = 108/35 ≈ 3.085, nope.

k = 16: 3*256/(256 - 4) = 768/252 = 64/21 ≈ 3.047, nope.

So, as k increases, a approaches 3.

Therefore, no other integer values of a except when k = 4.

Therefore, the answer is 4.

Therefore, I think the answer is 4.

But maybe the problem is expecting a different answer? Let me check with a = 5.

Wait, if a = 5, then the sum is 10, which is not a perfect square.

But maybe there is a different interpretation of the series? For example, maybe the series is finite? But the problem says "geometric series", which is usually infinite.

Alternatively, maybe the series is written as 4 + 12/a + 36/a² + ... and the ... implies that it's finite? But that's not standard.

Alternatively, maybe the problem is in another language, and "perfect square" refers to something else?

Alternatively, maybe I need to consider that the sum is a perfect square in the sense of being a square of an integer, but allowing a to be such that the sum is a square even if it's not integer? But that doesn't make sense.

Alternatively, maybe there's a mistake in the problem statement?

Alternatively, maybe I need to consider that the common ratio is negative? But since all terms are positive, the common ratio must be positive.

Therefore, I think my answer is correct.

Therefore, the smallest possible value of a is 4.

**Final Answer**
The smallest possible value of $ a $ is \boxed{4}.

To solve the problem, we are given the infinite geometric series:

$$
4 + \frac{12}{a} + \frac{36}{a^2} + \cdots
$$

We are to find the **smallest positive integer** $ a $ such that the **sum of the series is a perfect square**.

---

### Step 1: Identify the First Term and Common Ratio

- The **first term** $ a_1 = 4 $
- The **second term** is $ \frac{12}{a} $, so the **common ratio** $ r $ is:

$$
r = \frac{\frac{12}{a}}{4} = \frac{3}{a}
$$

---

### Step 2: Use the Formula for the Sum of an Infinite Geometric Series

The sum $ S $ of an infinite geometric series is given by:

$$
S = \frac{a_1}{1 - r}, \quad \text{if } |r| < 1
$$

Substituting the known values:

$$
S = \frac{4}{1 - \frac{3}{a}} = \frac{4}{\frac{a - 3}{a}} = \frac{4a}{a - 3}
$$

So, the sum of the series is:

$$
S = \frac{4a}{a - 3}
$$

We are told that this sum must be a **perfect square**. Let’s denote:

$$
\frac{4a}{a - 3} = k^2 \quad \text{for some integer } k
$$

---

### Step 3: Solve for $ a $ in Terms of $ k $

Multiply both sides by $ a - 3 $:

$$
4a = k^2(a - 3)
$$

Expanding and rearranging:

$$
4a = k^2 a - 3k^2 \Rightarrow 4a - k^2 a = -3k^2 \Rightarrow a(4 - k^2) = -3k^2
$$

Solving for $ a $:

$$
a = \frac{3k^2}{k^2 - 4}
$$

---

### Step 4: Ensure $ a $ is a Positive Integer

For $ a $ to be a **positive integer**, the denominator $ k^2 - 4 $ must divide the numerator $ 3k^2 $, and the result must be a positive integer.

Let’s analyze the expression:

$$
a = \frac{3k^2}{k^2 - 4}
$$

We can rewrite this as:

$$
a = 3 + \frac{12}{k^2 - 4}
$$

So, for $ a $ to be an integer, $ \frac{12}{k^2 - 4} $ must be an integer. That means $ k^2 - 4 $ must be a **positive divisor of 12**.

---

### Step 5: Find Valid Values of $ k $

The positive divisors of 12 are:

$$
1, 2, 3, 4, 6, 12
$$

So, $ k^2 - 4 \in \{1, 2, 3, 4, 6, 12\} \Rightarrow k^2 \in \{5, 6, 7, 8, 10, 16\} $

Now, among these, only $ k^2 = 16 $ is a **perfect square**. So:

$$
k = 4 \Rightarrow k^2 = 16
$$

Then:

$$
a = \frac{3k^2}{k^2 - 4} = \frac{3 \cdot 16}{16 - 4} = \frac{48}{12} = 4
$$

---

### Step 6: Verify the Result

Let’s check the sum when $ a = 4 $:

$$
S = \frac{4a}{a - 3} = \frac{4 \cdot 4}{4 - 3} = \frac{16}{1} = 16
$$

Which is a perfect square: $ 16 = 4^2 $

Also, the common ratio is $ r = \frac{3}{a} = \frac{3}{4} $, and since $ |r| < 1 $, the series converges.

---

### Final Answer

The smallest possible value of $ a $ such that the sum of the series is a perfect square is:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive integer value of 'a' such that the sum of the geometric series 4 + 12/a + 36/a² + ... is a perfect square. Let me start by recalling the formula for the sum of an infinite geometric series.

The formula is S = a₁ / (1 - r), where a₁ is the first term and r is the common ratio, provided that |r| < 1.

First, let me identify the first term and the common ratio of the given series. The first term, a₁, is clearly 4. Now, the second term is 12/a. To find the common ratio r, I can divide the second term by the first term. So, r = (12/a) / 4 = (12/a) * (1/4) = 3/a.

Let me check if this ratio makes sense. The third term should be the second term multiplied by r. Let's see: second term is 12/a, multiplied by r = 3/a gives (12/a)*(3/a) = 36/a², which matches the third term given. So that's correct.

So the common ratio r is 3/a. Now, for the series to converge, we need |r| < 1. Since a is a positive integer, and the terms are positive, we can drop the absolute value and just require that r < 1. Therefore, 3/a < 1. Solving for a, we get a > 3. So a must be an integer greater than 3.

Now, the sum S of the series is given by S = a₁ / (1 - r) = 4 / (1 - 3/a). Let me simplify that denominator.

First, 1 - 3/a is equal to (a - 3)/a. Therefore, S = 4 / [(a - 3)/a] = 4 * [a / (a - 3)] = (4a)/(a - 3).

So the sum S is (4a)/(a - 3). The problem states that this sum must be a perfect square. Therefore, we need (4a)/(a - 3) to be a perfect square.

Our task is to find the smallest positive integer a > 3 such that (4a)/(a - 3) is a perfect square.

Let me denote S = (4a)/(a - 3) = k², where k is a positive integer.

So, we have the equation:

(4a)/(a - 3) = k².

Let me solve this equation for a in terms of k.

Multiply both sides by (a - 3):

4a = k²(a - 3)

Expanding the right side:

4a = k²a - 3k²

Bring all terms to one side:

4a - k²a + 3k² = 0

Factor out a:

a(4 - k²) + 3k² = 0

Then:

a(4 - k²) = -3k²

Multiply both sides by -1:

a(k² - 4) = 3k²

Therefore:

a = (3k²)/(k² - 4)

So, a must be equal to 3k² divided by (k² - 4). Since a must be a positive integer, we need (k² - 4) to divide 3k², and also k² - 4 must be positive because a is positive.

So, k² - 4 > 0 => k² > 4 => k > 2. Since k is a positive integer, k ≥ 3.

Therefore, k is an integer greater than or equal to 3.

Our goal is to find the smallest a, so we need to find the smallest k ≥ 3 such that (k² - 4) divides 3k², and then compute a from that.

Alternatively, since a must be an integer, (k² - 4) must divide 3k². Let me think about how to approach this.

Let me write a = 3k² / (k² - 4). Let me try to simplify this expression.

Let me perform the division:

a = 3k² / (k² - 4) = 3 + [12 / (k² - 4)]

Wait, let me check that. Let me do polynomial division or manipulate the expression:

Let me write 3k² as 3(k² - 4) + 12.

Yes:

3(k² - 4) = 3k² - 12. Therefore, 3k² = 3(k² - 4) + 12.

Therefore, substituting back into a:

a = [3(k² - 4) + 12]/(k² - 4) = 3 + 12/(k² - 4)

Therefore, a = 3 + 12/(k² - 4).

Since a must be an integer, 12/(k² - 4) must be an integer. Therefore, (k² - 4) must be a positive divisor of 12.

So, k² - 4 divides 12. Also, since k² - 4 is positive, as established earlier, k² - 4 is a positive divisor of 12.

Therefore, let me list all positive divisors of 12.

The positive divisors of 12 are: 1, 2, 3, 4, 6, 12.

Therefore, k² - 4 must be one of these numbers.

So, k² - 4 ∈ {1, 2, 3, 4, 6, 12}

Therefore, k² ∈ {5, 6, 7, 8, 10, 16}

But k is an integer ≥ 3, so let's check which of these values of k² are perfect squares.

Looking at the numbers:

5: not a perfect square.

6: not a perfect square.

7: not a perfect square.

8: not a perfect square.

10: not a perfect square.

16: yes, 4² = 16.

So, only k² = 16 is a perfect square in that list. Therefore, k² - 4 = 12, so k² = 16, so k = 4.

Wait, but that seems like only one possible value? But maybe I missed something here. Let me check again.

Wait, the divisors of 12 are 1, 2, 3, 4, 6, 12. So k² - 4 must be one of these. Therefore, k² must be 5, 6, 7, 8, 10, or 16.

But only 16 is a perfect square. Therefore, k = 4.

But then, if k = 4, then a = 3 + 12/(k² - 4) = 3 + 12/12 = 3 + 1 = 4.

But wait, earlier we had that a must be greater than 3. So a = 4 is allowed.

But let me check if this works. Let me compute the sum S when a = 4.

Sum S = 4a/(a - 3) = 4*4/(4 - 3) = 16/1 = 16. Which is 4², a perfect square.

So that works. But the question says "the smallest possible value of a where a is a positive integer". But a = 4 is possible. However, wait, the problem says "the geometric series 4 + 12/a + 36/a² + ...". If a = 4, then the series is 4 + 12/4 + 36/16 + ... which is 4 + 3 + 2.25 + ... and the sum is 16.

But the problem says "the smallest possible value of a". But maybe there are other values of k where k² - 4 divides 12, but k² is not necessarily in the list? Wait, no. Because k² - 4 must divide 12, so k² - 4 is a positive divisor of 12, so k² must be 4 + d, where d is a positive divisor of 12.

But maybe there are other possibilities? Wait, but if k² - 4 divides 12, then k² - 4 must be a positive integer divisor of 12. Therefore, the possible values of k² - 4 are 1, 2, 3, 4, 6, 12. Therefore, k² is 5, 6, 7, 8, 10, 16.

But only 16 is a perfect square. Therefore, only k = 4. Therefore, only a = 4.

But that seems too easy. However, the problem says "the smallest possible value of a where a is a positive integer". If a = 4 is possible, then that would be the answer. But maybe I made a mistake here? Let me check again.

Wait, let me check if there are other divisors. Wait, maybe negative divisors? But since k² - 4 is positive, we don't need to consider negative divisors.

Alternatively, perhaps my approach is missing something? Let me think again.

We had that a = 3k²/(k² - 4). So, for a to be integer, (k² - 4) must divide 3k².

But maybe there are other ways for (k² - 4) to divide 3k² without necessarily k² - 4 dividing 12? Wait, no. Because earlier, I rewrote a as 3 + 12/(k² - 4). Therefore, 12/(k² - 4) must be integer. Therefore, k² - 4 divides 12. Therefore, my previous reasoning holds.

Therefore, only k = 4 gives a valid solution. Therefore, a = 4.

But let me check if there are other possibilities. For example, maybe k² - 4 divides 3k², but not necessarily divides 12? Wait, but in the expression a = 3 + 12/(k² - 4), the term 12/(k² - 4) must be integer. Therefore, k² - 4 divides 12. Therefore, my previous conclusion is correct.

But then, if a = 4 is possible, then why does the problem ask for the smallest possible value? Maybe there is a mistake here? Let me check with a = 4.

Original series: 4 + 12/4 + 36/16 + ...

Which is 4 + 3 + 2.25 + 1.6875 + ...

Sum is 16, which is 4². So that's correct.

But maybe there is a smaller a? Wait, a must be greater than 3. The possible a's are integers starting from 4. So 4 is the smallest possible.

But the problem says "the smallest possible value of a where a is a positive integer". So 4 is the answer? But maybe I made a mistake in assuming that k² - 4 divides 12? Let me check again.

Wait, let me think again.

We had:

a = 3k²/(k² - 4).

So, for a to be integer, (k² - 4) must divide 3k².

But perhaps there are other values of k where (k² - 4) divides 3k², even if (k² - 4) does not divide 12?

Wait, but earlier, I rewrote a as 3 + 12/(k² - 4). Therefore, 12/(k² - 4) must be integer. Therefore, k² - 4 divides 12. Therefore, my previous reasoning is correct.

But let me check with k = 5.

If k = 5, then k² = 25. Then, k² - 4 = 21. Then, a = 3*(25)/21 = 75/21 = 25/7 ≈ 3.57, which is not integer.

Similarly, k = 3. Let me check k = 3.

k = 3, then k² = 9. Then, k² - 4 = 5. Then, a = 3*9 / 5 = 27/5 = 5.4, not integer.

k = 4, as before, gives a = 4.

k = 5, as above, gives non-integer.

k = 6: k² = 36, k² - 4 = 32. Then, a = 3*36 / 32 = 108/32 = 27/8, not integer.

k = 2: but k must be ≥ 3.

k = 1: no.

k = 0: no.

So, only k = 4 gives integer a. Therefore, a = 4.

But the problem says "the smallest possible value of a where a is a positive integer". So 4 is the answer?

But maybe there's a mistake here. Let me check again.

Wait, let me think if there are other possibilities where k² - 4 divides 3k², but not necessarily divides 12.

Wait, for example, suppose that k² - 4 divides 3k². Let me write that as:

Let d = k² - 4. Then, d divides 3k². Also, since d = k² - 4, then k² = d + 4. Therefore, substituting into 3k², we get 3(d + 4) = 3d + 12. Therefore, d divides 3d + 12. Which implies that d divides 12. Because if d divides 3d + 12, then d divides (3d + 12) - 3*d = 12.

Therefore, d divides 12. Therefore, my previous conclusion is correct. Therefore, d must be a positive divisor of 12. Therefore, k² - 4 is a positive divisor of 12. Therefore, only possible values of k² - 4 are 1, 2, 3, 4, 6, 12. Therefore, only k² = 5, 6, 7, 8, 10, 16.

Only k² = 16 is a perfect square, so k = 4. Therefore, a = 4.

Therefore, the answer is 4? But maybe the problem is expecting a different answer? Let me check again.

Wait, let me check if there are other divisors. For example, if d divides 12, but d could be negative? But since d = k² - 4, and k is integer ≥ 3, then d is positive. Therefore, no.

Alternatively, maybe I made a mistake in the formula for the sum? Let me check again.

Original series: 4 + 12/a + 36/a² + ...

First term a₁ = 4. Common ratio r = (12/a)/4 = 3/a.

Sum S = a₁ / (1 - r) = 4 / (1 - 3/a) = 4 / [(a - 3)/a] = 4a / (a - 3).

Yes, that's correct.

So S = 4a/(a - 3).

So if a = 4, S = 16/1 = 16, which is 4².

But maybe there is a smaller a? Wait, a must be greater than 3. So the next possible integer is 4. Therefore, 4 is the minimal.

But maybe the problem is in another interpretation? Let me check the problem again.

"Consider the geometric series 4 + 12/a + 36/a² + ... . If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?"

So, the problem says "the smallest possible value of a where a is a positive integer". So if a = 4 works, then that's the answer.

But maybe there is a mistake here? Let me check with a = 5.

If a = 5, then S = 4*5/(5 - 3) = 20/2 = 10. 10 is not a perfect square.

a = 6: 4*6/(6 - 3) = 24/3 = 8, not a perfect square.

a = 7: 4*7/(7 - 3) = 28/4 = 7, not a perfect square.

a = 8: 4*8/(8 - 3) = 32/5 = 6.4, not integer.

a = 9: 4*9/(9 - 3) = 36/6 = 6, not a perfect square.

a = 10: 4*10/(10 - 3) = 40/7 ≈ 5.71, not integer.

a = 12: 4*12/(12 - 3) = 48/9 = 16/3 ≈ 5.33, not integer.

a = 16: 4*16/(16 - 3) = 64/13 ≈ 4.92, not integer.

Wait, but according to our previous reasoning, only a = 4 works.

But let me check for k = something else. Wait, maybe there are other k where k² - 4 divides 12? For example, if k² - 4 is 1, 2, 3, 4, 6, 12.

But k² - 4 = 1 => k² = 5, which is not a square.

k² - 4 = 2 => k² = 6, not square.

k² - 4 = 3 => k² = 7, nope.

k² - 4 = 4 => k² = 8, nope.

k² - 4 = 6 => k² = 10, nope.

k² - 4 = 12 => k² = 16, which is 4².

Therefore, only k = 4.

Therefore, only a = 4.

But maybe there's another approach where a is not necessarily derived from k? Let me think.

Suppose that S = 4a/(a - 3) is a perfect square. Let me denote S = m², where m is a positive integer.

So, 4a/(a - 3) = m².

Let me rearrange this equation:

4a = m²(a - 3)

=> 4a = m²a - 3m²

=> 4a - m²a = -3m²

=> a(4 - m²) = -3m²

=> a = (3m²)/(m² - 4)

Which is the same as before.

So, same result. Therefore, same conclusion.

Therefore, the only possible value is m = 4, leading to a = 4.

But maybe m is not an integer? Wait, no, because S is a perfect square, so m must be integer.

Alternatively, maybe m is rational? But since S is a perfect square, and S is equal to 4a/(a - 3), which is rational because a is integer. Therefore, if S is a perfect square, then it must be a square of a rational number. However, since the problem says "a perfect square", it's likely referring to an integer perfect square.

But let me check. Suppose that S is a perfect square in the sense of rational numbers. For example, maybe S is (p/q)² where p and q are integers. However, the problem says "the sum is a perfect square", and in math competitions, usually, unless specified otherwise, perfect square refers to integer perfect squares.

Therefore, assuming that S must be an integer perfect square. Therefore, our previous reasoning holds.

Therefore, the answer is a = 4.

But let me check again with a = 4.

Sum is 16, which is 4². So that's correct.

But maybe the problem is expecting a different answer? Maybe I made a mistake in the initial steps?

Wait, let me check the problem statement again.

"Consider the geometric series 4 + 12/a + 36/a² + ... . If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?"

Wait, the series is 4 + 12/a + 36/a² + ... So the first term is 4, second term is 12/a, third term is 36/a², etc.

So the ratio between the second term and the first term is (12/a)/4 = 3/a. Then the ratio between the third term and the second term is (36/a²)/(12/a) = 3/a. So the common ratio is indeed 3/a.

Therefore, the sum is 4/(1 - 3/a) = 4a/(a - 3).

So that's correct.

Therefore, the answer is 4.

But maybe the problem is from a source where the answer is different? Wait, maybe I need to check if there are other values of a where 4a/(a - 3) is a perfect square.

Let me suppose that 4a/(a - 3) is a perfect square. Let me denote that as m².

So, 4a/(a - 3) = m².

Let me rearrange this equation:

Multiply both sides by (a - 3):

4a = m²(a - 3)

=> 4a = m²a - 3m²

=> 4a - m²a = -3m²

=> a(4 - m²) = -3m²

=> a = (3m²)/(m² - 4)

So, same as before.

Therefore, for a to be positive integer, m² - 4 must divide 3m² and m² - 4 must be positive.

So m² - 4 divides 3m².

Let me think of m² - 4 divides 3m².

Let me write m² - 4 divides 3m².

Which implies that m² - 4 divides 3m² - 3*(m² - 4) = 3m² - 3m² + 12 = 12.

Therefore, m² - 4 divides 12.

Therefore, m² - 4 is a positive divisor of 12.

Therefore, same conclusion as before.

Therefore, m² - 4 ∈ {1, 2, 3, 4, 6, 12}

Therefore, m² ∈ {5, 6, 7, 8, 10, 16}

Only m² = 16 is a perfect square. Therefore, m = 4.

Therefore, a = 3*(16)/(16 - 4) = 48/12 = 4.

Therefore, the answer is 4.

But maybe the problem is expecting a different answer? Let me check with m = something else.

Wait, suppose m² - 4 divides 12, but m² - 4 is negative? But since m² - 4 must be positive, as we saw earlier.

Alternatively, maybe m is not integer? But if m is rational, then S is a square of rational number. But the problem says "perfect square", which is ambiguous. However, in most cases, unless specified, perfect square refers to integer.

But let me check if there are other possibilities where S is a square of a rational number.

Suppose that S = (p/q)², where p and q are integers with no common factors. Then, 4a/(a - 3) = p²/q².

Cross multiplying: 4a q² = p²(a - 3).

But this seems more complicated. However, since a is integer, maybe there are solutions where S is a rational square but not integer. However, the problem says "the sum is a perfect square". If they mean a square of a rational number, then there could be more solutions.

But given that the problem is likely from a competition or textbook problem, and given that the answer is expected to be an integer, and the sum is 16 when a = 4, which is an integer perfect square, I think that the answer is 4.

But maybe there is a mistake here. Let me check with a = 5.

If a = 5, then S = 4*5/(5 - 3) = 20/2 = 10. 10 is not a perfect square.

a = 6: 24/3 = 8, not square.

a = 7: 28/4 = 7, not square.

a = 8: 32/5 = 6.4, not integer.

a = 9: 36/6 = 6, not square.

a = 10: 40/7 ≈ 5.71, not integer.

a = 12: 48/9 = 16/3 ≈ 5.33, not integer.

a = 16: 64/13 ≈ 4.92, not integer.

a = 20: 80/17 ≈ 4.7, not integer.

a = 24: 96/21 = 32/7 ≈ 4.57, not integer.

a = 3: but a must be greater than 3.

a = 4: 16, which is square.

So, seems like a = 4 is the only possible.

But maybe there is another way where m² - 4 divides 12, but m is not integer? For example, if m is rational.

But if m is rational, then m² is rational. However, since S = 4a/(a - 3) must be rational, which it is, but the problem says "perfect square". If they mean square of rational number, then there could be more solutions.

But let me think. Suppose that S is a square of a rational number. Then, 4a/(a - 3) = (p/q)².

Cross multiplying: 4a q² = p²(a - 3).

But this equation may have solutions for other a. However, since the problem says "the sum is a perfect square", and given that the answer is likely to be an integer, and given that a = 4 works, maybe that's the answer.

Alternatively, maybe the problem is from a source where the answer is different. Wait, let me check if there is a mistake in my calculation.

Wait, let me check for a = 5.

If a = 5, then the sum is 10. Is 10 a perfect square? No.

a = 6: 8, nope.

a = 7: 7, nope.

a = 8: 32/5, nope.

a = 9: 6, nope.

a = 10: 40/7, nope.

a = 12: 16/3, nope.

a = 15: 4*15/(15 - 3) = 60/12 = 5, not square.

a = 16: 64/13, nope.

a = 20: 80/17, nope.

a = 24: 96/21 = 32/7, nope.

a = 3: invalid.

a = 4: 16, which is square.

So, seems like only a = 4.

But maybe there is another value of a where 4a/(a - 3) is a perfect square. Let me think algebraically.

Let me set 4a/(a - 3) = m².

Let me write this as 4a = m²(a - 3).

Let me rearrange:

4a = m²a - 3m²

=> 4a - m²a = -3m²

=> a(4 - m²) = -3m²

=> a = (3m²)/(m² - 4)

So, for a to be integer, m² - 4 must divide 3m².

Let me think of m² - 4 divides 3m².

Let me write m² - 4 divides 3m².

Let me denote d = m² - 4. Then, d divides 3m².

But since d = m² - 4, then m² = d + 4.

Therefore, substituting into 3m², we get 3(d + 4) = 3d + 12.

Therefore, d divides 3d + 12 => d divides 12.

Therefore, d is a positive divisor of 12.

Therefore, same conclusion.

Therefore, only possible d is 1, 2, 3, 4, 6, 12.

Therefore, m² = d + 4.

So, m² = 5, 6, 7, 8, 10, 16.

Only m² = 16 is a perfect square.

Therefore, m = 4, leading to a = 3*(16)/(16 - 4) = 48/12 = 4.

Therefore, the answer is 4.

Therefore, I think the answer is 4.

But maybe the problem is in another interpretation? For example, maybe the series is finite? But the problem says "geometric series", which usually implies infinite unless specified otherwise.

Alternatively, maybe the problem is from a source where the answer is different. But given the calculations, I think 4 is correct.

Alternatively, maybe I made a mistake in the initial formula? Let me check again.

Sum of infinite geometric series is a₁/(1 - r).

First term a₁ = 4.

Common ratio r = 3/a.

Therefore, sum S = 4 / (1 - 3/a) = 4 / [(a - 3)/a] = 4a/(a - 3).

Yes, that's correct.

Therefore, I think the answer is 4.

But to be thorough, let me check if there are other values of a where 4a/(a - 3) is a perfect square.

Suppose that 4a/(a - 3) = k².

Let me rearrange this equation:

Let me write it as 4a = k²(a - 3)

=> 4a = k²a - 3k²

=> 4a - k²a = -3k²

=> a(4 - k²) = -3k²

=> a = 3k²/(k² - 4)

So, for a to be integer, k² - 4 must divide 3k².

Let me think of k as integer ≥ 3.

Let me try k = 5:

a = 3*(25)/(25 - 4) = 75/21 = 25/7 ≈ 3.57, not integer.

k = 6: 3*36/(36 - 4) = 108/32 = 27/8 ≈ 3.375, not integer.

k = 7: 3*49/(49 - 4) = 147/45 = 49/15 ≈ 3.266..., nope.

k = 8: 3*64/(64 - 4) = 192/60 = 16/5 = 3.2, nope.

k = 9: 3*81/(81 - 4) = 243/77 ≈ 3.155, nope.

k = 10: 3*100/(100 - 4) = 300/96 = 25/8 = 3.125, nope.

k = 12: 3*144/(144 - 4) = 432/140 = 108/35 ≈ 3.085, nope.

k = 16: 3*256/(256 - 4) = 768/252 = 64/21 ≈ 3.047, nope.

So, as k increases, a approaches 3.

Therefore, no other integer values of a except when k = 4.

Therefore, the answer is 4.

Therefore, I think the answer is 4.

But maybe the problem is expecting a different answer? Let me check with a = 5.

Wait, if a = 5, then the sum is 10, which is not a perfect square.

But maybe there is a different interpretation of the series? For example, maybe the series is finite? But the problem says "geometric series", which is usually infinite.

Alternatively, maybe the series is written as 4 + 12/a + 36/a² + ... and the ... implies that it's finite? But that's not standard.

Alternatively, maybe the problem is in another language, and "perfect square" refers to something else?

Alternatively, maybe I need to consider that the sum is a perfect square in the sense of being a square of an integer, but allowing a to be such that the sum is a square even if it's not integer? But that doesn't make sense.

Alternatively, maybe there's a mistake in the problem statement?

Alternatively, maybe I need to consider that the common ratio is negative? But since all terms are positive, the common ratio must be positive.

Therefore, I think my answer is correct.

Therefore, the smallest possible value of a is 4.

**Final Answer**
The smallest possible value of $ a $ is \boxed{4}.

To solve the problem, we are given the infinite geometric series:

$$
4 + \frac{12}{a} + \frac{36}{a^2} + \cdots
$$

We are to find the **smallest positive integer** $ a $ such that the **sum of the series is a perfect square**.

---

### Step 1: Identify the First Term and Common Ratio

- The **first term** $ a_1 = 4 $
- The **second term** is $ \frac{12}{a} $, so the **common ratio** $ r $ is:

$$
r = \frac{\frac{12}{a}}{4} = \frac{3}{a}
$$

---

### Step 2: Use the Formula for the Sum of an Infinite Geometric Series

The sum $ S $ of an infinite geometric series is given by:

$$
S = \frac{a_1}{1 - r}, \quad \text{if } |r| < 1
$$

Substituting the known values:

$$
S = \frac{4}{1 - \frac{3}{a}} = \frac{4}{\frac{a - 3}{a}} = \frac{4a}{a - 3}
$$

So, the sum of the series is:

$$
S = \frac{4a}{a - 3}
$$

We are told that this sum must be a **perfect square**. Let’s denote:

$$
\frac{4a}{a - 3} = k^2 \quad \text{for some integer } k
$$

---

### Step 3: Solve for $ a $ in Terms of $ k $

Multiply both sides by $ a - 3 $:

$$
4a = k^2(a - 3)
$$

Expanding and rearranging:

$$
4a = k^2 a - 3k^2 \Rightarrow 4a - k^2 a = -3k^2 \Rightarrow a(4 - k^2) = -3k^2
$$

Solving for $ a $:

$$
a = \frac{3k^2}{k^2 - 4}
$$

---

### Step 4: Ensure $ a $ is a Positive Integer

For $ a $ to be a **positive integer**, the denominator $ k^2 - 4 $ must divide the numerator $ 3k^2 $, and the result must be a positive integer.

Let’s analyze the expression:

$$
a = \frac{3k^2}{k^2 - 4}
$$

We can rewrite this as:

$$
a = 3 + \frac{12}{k^2 - 4}
$$

So, for $ a $ to be an integer, $ \frac{12}{k^2 - 4} $ must be an integer. That means $ k^2 - 4 $ must be a **positive divisor of 12**.

---

### Step 5: Find Valid Values of $ k $

The positive divisors of 12 are:

$$
1, 2, 3, 4, 6, 12
$$

So, $ k^2 - 4 \in \{1, 2, 3, 4, 6, 12\} \Rightarrow k^2 \in \{5, 6, 7, 8, 10, 16\} $

Now, among these, only $ k^2 = 16 $ is a **perfect square**. So:

$$
k = 4 \Rightarrow k^2 = 16
$$

Then:

$$
a = \frac{3k^2}{k^2 - 4} = \frac{3 \cdot 16}{16 - 4} = \frac{48}{12} = 4
$$

---

### Step 6: Verify the Result

Let’s check the sum when $ a = 4 $:

$$
S = \frac{4a}{a - 3} = \frac{4 \cdot 4}{4 - 3} = \frac{16}{1} = 16
$$

Which is a perfect square: $ 16 = 4^2 $

Also, the common ratio is $ r = \frac{3}{a} = \frac{3}{4} $, and since $ |r| < 1 $, the series converges.

---

### Final Answer

The smallest possible value of $ a $ such that the sum of the series is a perfect square is:

$$
\boxed{4}
$$

Solution:
We use the formula $\left(\frac{\text{first term}}{1-(\text{common ratio})}\right)$ for the sum of a geometric series to get the sum $\left(\frac{4}{1-\frac{3}{a}}\right)=\frac{4}{\frac{a-3}{a}}=\frac{4a}{a-3}$. We want $\frac{4a}{a-3}$ to be a perfect square $b^2$, where $b$ is a positive integer. So we have $4a=b^2(a-3)$ and start trying values for $b$ until we get a positive integer $a$.
If $b=1$, then $4a=a-3$, but that means $a=-1$.
If $b=2$, then $4a=4(a-3)\qquad\Rightarrow 0=-12$.
If $b=3$, then $4a=9(a-3)\qquad\Rightarrow -5a=-27$, which doesn't yield an integer value for $a$.
If $b=4$, then $4a=16(a-3)\qquad\Rightarrow -12a=-48$, so $a=\boxed{4}$, which is a positive integer.

OR

For an infinite geometric series to converge, the common ratio must be between $-1$ and $1$. Thus $\frac{3}{a}$ must be less than 1, which means $a$ is greater than 3. We try $a=4$ and get that $\left(\frac{4}{1-\frac{3}{4}}\right)=\left(\frac{4}{\frac{1}{4}}\right)=4\cdot4=16$, which is a perfect square.