Explanation:
Okay, so I need to find the number of quadratic equations of the form $x^2 + ax + b = 0$ such that whenever $c$ is a root, $c^2 - 2$ is also a root. Hmm, let me think about how to approach this.

First, let me recall that a quadratic equation has two roots (counting multiplicity). Let me denote the roots as $r_1$ and $r_2$. According to Vieta's formulas, we have:

$r_1 + r_2 = -a$ and $r_1 r_2 = b$.

Now, the condition given is that if $c$ is a root, then $c^2 - 2$ is also a root. So, for each root $c$, $c^2 - 2$ must be another root. Since there are only two roots, this condition must hold for both roots. Let me consider the possibilities.

Case 1: The quadratic has a repeated root. That is, $r_1 = r_2$. Then, if $c = r_1$, then $c^2 - 2$ must also be a root. But since there's only one root, $c^2 - 2$ must equal $c$. So, solving $c^2 - 2 = c$, which gives $c^2 - c - 2 = 0$. Solving this quadratic equation:

$c = [1 \pm \sqrt{1 + 8}]/2 = [1 \pm 3]/2$, so $c = 2$ or $c = -1$.

Therefore, if the quadratic has a repeated root, it must be either 2 or -1. Let me check if these satisfy the condition.

If the root is 2, then $c^2 - 2 = 4 - 2 = 2$, which is the same root. So that works. Similarly, if the root is -1, then $c^2 - 2 = 1 - 2 = -1$, which is also the same root. So both 2 and -1 are valid repeated roots. Therefore, there are two quadratics in this case: $(x - 2)^2 = x^2 - 4x + 4$ and $(x + 1)^2 = x^2 + 2x + 1$. So that's two equations.

Case 2: The quadratic has two distinct roots. Let me denote them as $r$ and $s$. Then, according to the condition, if $r$ is a root, then $r^2 - 2$ must be a root. Similarly, if $s$ is a root, then $s^2 - 2$ must be a root. Since there are only two roots, this implies that $r^2 - 2$ must be either $r$ or $s$, and similarly for $s^2 - 2$.

So, let me consider the possibilities for each root:

For root $r$, $r^2 - 2$ is either $r$ or $s$.

Similarly, for root $s$, $s^2 - 2$ is either $r$ or $s$.

Let me consider different subcases.

Subcase 2a: Suppose $r^2 - 2 = r$ and $s^2 - 2 = s$. Then, this would imply that both roots satisfy $x^2 - x - 2 = 0$, which factors as $(x - 2)(x + 1) = 0$. Therefore, the roots would be 2 and -1. But then, this is the case where the quadratic has roots 2 and -1. However, in this case, the quadratic would be $(x - 2)(x + 1) = x^2 - x - 2$. But wait, this is a quadratic with distinct roots. However, we need to check if this satisfies the original condition. Let me check:

If $c = 2$, then $c^2 - 2 = 4 - 2 = 2$, which is a root. If $c = -1$, then $c^2 - 2 = 1 - 2 = -1$, which is also a root. So this quadratic does satisfy the condition. However, this quadratic is different from the ones in Case 1. Wait, but in Case 1, we had quadratics with repeated roots. So this is a separate case. Therefore, this quadratic is valid. However, note that this quadratic is different from the ones in Case 1. So this is another quadratic. But wait, but in this case, the roots are 2 and -1, so the quadratic is $x^2 - x - 2$. However, in Case 1, we had quadratics with repeated roots 2 and -1. So this is a different quadratic. Therefore, this is another case. However, let me check if there are other possibilities.

Subcase 2b: Suppose that for one root, $r^2 - 2 = s$, and for the other root, $s^2 - 2 = r$. So, this would mean that applying the transformation $c \mapsto c^2 - 2$ swaps the roots. So, we have:

$r^2 - 2 = s$ and $s^2 - 2 = r$.

So, we can set up these equations:

From the first equation: $s = r^2 - 2$.

Substitute into the second equation: $(r^2 - 2)^2 - 2 = r$.

Let me expand that:

$(r^4 - 4r^2 + 4) - 2 = r$

Simplify:

$r^4 - 4r^2 + 2 = r$

Bring all terms to one side:

$r^4 - 4r^2 - r + 2 = 0$

Hmm, quartic equation. Let me try to factor this. Maybe there are rational roots. Let me apply Rational Root Theorem. Possible rational roots are ±1, ±2.

Testing r = 1: 1 - 4 - 1 + 2 = -2 ≠ 0.

Testing r = -1: 1 - 4 + 1 + 2 = 0. Oh, r = -1 is a root.

So, we can factor (r + 1) out of the quartic.

Let me perform polynomial division or use synthetic division.

Divide $r^4 - 4r^2 - r + 2$ by (r + 1).

Using synthetic division for root -1:

Coefficients: 1 (r^4), 0 (r^3), -4 (r^2), -1 (r), 2 (constant).

Bring down the 1.

Multiply by -1: -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: 1. Add to next coefficient: -4 + 1 = -3.

Multiply by -1: 3. Add to next coefficient: -1 + 3 = 2.

Multiply by -1: -2. Add to last coefficient: 2 + (-2) = 0. Perfect.

So the quartic factors as (r + 1)(r^3 - r^2 - 3r + 2).

Now, let's factor the cubic: $r^3 - r^2 - 3r + 2$.

Again, try rational roots: ±1, ±2.

Testing r = 1: 1 - 1 - 3 + 2 = -1 ≠ 0.

Testing r = 2: 8 - 4 - 6 + 2 = 0. Yes, r = 2 is a root.

Divide the cubic by (r - 2):

Using synthetic division:

Coefficients: 1 (r^3), -1 (r^2), -3 (r), 2 (constant).

Root at 2:

Bring down 1.

Multiply by 2: 2. Add to -1: 1.

Multiply by 2: 2. Add to -3: -1.

Multiply by 2: -2. Add to 2: 0. Perfect.

So the cubic factors as (r - 2)(r^2 + r - 1).

Therefore, the quartic factors as (r + 1)(r - 2)(r^2 + r - 1).

Therefore, the roots of the quartic equation are r = -1, r = 2, and roots of $r^2 + r - 1 = 0$, which are $r = [-1 \pm \sqrt{1 + 4}]/2 = [-1 \pm \sqrt{5}]/2$.

So, going back to our original equations, we had:

s = r^2 - 2, and s^2 - 2 = r.

So, for each root r of the quartic, we can find s. However, we need to check if these lead to valid quadratics.

First, let's consider r = -1. Then s = (-1)^2 - 2 = 1 - 2 = -1. Therefore, s = -1. So, this gives us the case where both roots are -1, which is the repeated root case we considered earlier. So this is already covered in Case 1.

Similarly, for r = 2: s = 2^2 - 2 = 4 - 2 = 2. Again, this gives s = 2, which is the repeated root case. So again, covered in Case 1.

Now, for the roots of $r^2 + r - 1 = 0$, which are $r = [-1 + \sqrt{5}]/2$ and $r = [-1 - \sqrt{5}]/2$. Let me denote them as $r_1$ and $r_2$.

So, for each of these roots, we can compute s = r^2 - 2. Let me compute s for $r_1$:

First, $r_1 = [-1 + \sqrt{5}]/2$. Let me compute $r_1^2$:

$r_1^2 = [(-1 + \sqrt{5})^2]/4 = [1 - 2\sqrt{5} + 5]/4 = [6 - 2\sqrt{5}]/4 = [3 - \sqrt{5}]/2$.

Then, $s = r_1^2 - 2 = [3 - \sqrt{5}]/2 - 2 = [3 - \sqrt{5} - 4]/2 = [-1 - \sqrt{5}]/2 = r_2$.

Similarly, for $r_2 = [-1 - \sqrt{5}]/2$, compute $r_2^2$:

$r_2^2 = [(-1 - \sqrt{5})^2]/4 = [1 + 2\sqrt{5} + 5]/4 = [6 + 2\sqrt{5}]/4 = [3 + \sqrt{5}]/2$.

Then, $s = r_2^2 - 2 = [3 + \sqrt{5}]/2 - 2 = [3 + \sqrt{5} - 4]/2 = [-1 + \sqrt{5}]/2 = r_1$.

Therefore, for these roots, we have that $s = r_2$ when $r = r_1$, and $s = r_1$ when $r = r_2$. Therefore, this gives us a pair of roots $r_1$ and $r_2$, which are distinct. Therefore, this corresponds to a quadratic equation with roots $r_1$ and $r_2$. Let me write down the quadratic equation.

Since the roots are $r_1$ and $r_2$, the quadratic is $x^2 - (r_1 + r_2)x + r_1 r_2$. Let me compute $r_1 + r_2$ and $r_1 r_2$.

From the quadratic equation $r^2 + r - 1 = 0$, we know that $r_1 + r_2 = -1$ and $r_1 r_2 = -1$. Therefore, the quadratic equation is $x^2 - (-1)x + (-1) = x^2 + x - 1$. Therefore, this quadratic is $x^2 + x - 1$. Let me check if this satisfies the original condition.

Take root $r_1 = [-1 + \sqrt{5}]/2$. Compute $c^2 - 2$:

As above, $c^2 - 2 = r_2$, which is the other root. Similarly for $r_2$, $c^2 - 2 = r_1$. Therefore, this quadratic satisfies the condition. Therefore, this is another valid quadratic equation.

So, in Subcase 2b, we have another quadratic equation $x^2 + x - 1$.

Wait, but hold on. Earlier, in Subcase 2a, we had the quadratic $x^2 - x - 2$ with roots 2 and -1. Let me check if that quadratic also satisfies the condition. Let me check if for each root, applying $c^2 - 2$ gives another root.

For root 2: $2^2 - 2 = 4 - 2 = 2$, which is the same root. For root -1: $(-1)^2 - 2 = 1 - 2 = -1$, same root. Therefore, this quadratic also satisfies the condition. However, in Subcase 2a, we considered the case where both roots satisfy $c^2 - 2 = c$, which led us to the roots 2 and -1. However, in this case, the quadratic is $x^2 - x - 2$, which is different from the quadratics in Case 1 (which had repeated roots). Therefore, this is another quadratic. So, in total, we have:

Case 1: two quadratics (repeated roots 2 and -1).

Case 2a: one quadratic (roots 2 and -1).

Case 2b: one quadratic (roots $r_1$ and $r_2$ from the equation $x^2 + x - 1 = 0$).

Wait, but hold on. Wait, in Case 2a, we considered the case where both roots satisfy $c^2 - 2 = c$, which gives roots 2 and -1. However, in Case 2b, we have another quadratic with roots $r_1$ and $r_2$ which are not 2 or -1. So, that's a separate quadratic. However, is there another case?

Wait, let me check if there are other possibilities. For example, in Subcase 2c: Suppose that for one root, $r^2 - 2 = r$, and for the other root, $s^2 - 2 = s$. Wait, that's exactly Subcase 2a. So that's already considered.

Alternatively, maybe there's another case where one root maps to itself, and the other maps to the first root? Wait, but if $r^2 - 2 = r$, then that's covered in Subcase 2a. If $s^2 - 2 = r$, then that's covered in Subcase 2b. So, perhaps those are the only possibilities.

Wait, but let me think again. Suppose that for one root, $r^2 - 2 = s$, and for the other root, $s^2 - 2 = s$. Then, that would be another case. Let me check.

Subcase 2c: Suppose $r^2 - 2 = s$ and $s^2 - 2 = s$. Then, from the second equation, $s^2 - s - 2 = 0$, which factors as $(s - 2)(s + 1) = 0$, so $s = 2$ or $s = -1$. Then, from the first equation, $r^2 - 2 = s$. Therefore, if $s = 2$, then $r^2 - 2 = 2$ implies $r^2 = 4$, so $r = 2$ or $r = -2$. Similarly, if $s = -1$, then $r^2 - 2 = -1$ implies $r^2 = 1$, so $r = 1$ or $r = -1$.

So, let's check these possibilities.

First, if $s = 2$, then $r$ can be 2 or -2. If $r = 2$, then the roots are 2 and 2, which is the repeated root case (already considered in Case 1). If $r = -2$, then the roots are -2 and 2. Let me check if this quadratic satisfies the condition. Let me check for each root:

For root 2: $2^2 - 2 = 2$, which is a root. For root -2: $(-2)^2 - 2 = 4 - 2 = 2$, which is a root. Therefore, this quadratic $x^2 - (2 + (-2))x + (2)(-2) = x^2 - 0x - 4 = x^2 - 4$. Wait, but hold on, if the roots are 2 and -2, then Vieta's formula gives sum = 0, product = -4. Therefore, the quadratic is $x^2 - 0x - 4 = x^2 - 4$. Let me check if this quadratic satisfies the condition. For root 2: 2^2 - 2 = 2, which is a root. For root -2: (-2)^2 - 2 = 4 - 2 = 2, which is a root. Therefore, yes, this quadratic satisfies the condition. Therefore, this is another quadratic equation. Similarly, if $s = -1$, then $r^2 - 2 = -1$ implies $r^2 = 1$, so $r = 1$ or $r = -1$. If $r = 1$, then the roots are 1 and -1. Let me check if this quadratic satisfies the condition. The quadratic would be $x^2 - (1 + (-1))x + (1)(-1) = x^2 - 0x -1 = x^2 - 1$. Let me check the roots: 1 and -1. For root 1: 1^2 - 2 = -1, which is a root. For root -1: (-1)^2 - 2 = 1 - 2 = -1, which is a root. Therefore, this quadratic also satisfies the condition. Similarly, if $r = -1$, then the roots are -1 and -1, which is the repeated root case already considered. Therefore, in this Subcase 2c, we have two new quadratics: $x^2 - 4$ and $x^2 - 1$.

Wait, so this seems like another case. Therefore, I need to check if these quadratics were already considered in previous cases. In Case 1, we had quadratics with repeated roots 2 and -1. Here, we have quadratics with distinct roots 2 and -2, and 1 and -1. Therefore, these are new quadratics. Therefore, this suggests that there are more quadratics to consider.

So, perhaps my initial analysis was incomplete. Let me try to organize all these cases.

So, let me recap:

Case 1: Repeated roots. We had quadratics with roots 2 and 2, and -1 and -1. So two quadratics.

Case 2: Distinct roots. Let me consider different possibilities:

Subcase 2a: Both roots satisfy $c^2 - 2 = c$. This gives roots 2 and -1, leading to quadratic $x^2 - x - 2$.

Subcase 2b: The roots are such that $r^2 - 2 = s$ and $s^2 - 2 = r$. This gives roots from $x^2 + x - 1 = 0$, leading to quadratic $x^2 + x - 1$.

Subcase 2c: One root satisfies $c^2 - 2 = c$ (i.e., 2 or -1) and the other root maps to the first root. For example, if one root is 2, then the other root is -2 (as in $x^2 - 4$), or if one root is -1, the other root is 1 (as in $x^2 - 1$). Wait, but in Subcase 2c, we considered when one root maps to itself (i.e., satisfies $c^2 - 2 = c$) and the other maps to the first root. Therefore, these quadratics $x^2 - 4$ and $x^2 - 1$ are valid.

But wait, are there more possibilities? For example, if one root maps to the other root, and the other root maps to itself? Wait, that's what Subcase 2c is. So, in addition to that, maybe there are other quadratics where neither root maps to itself, but they map to each other? Wait, that's Subcase 2b. So, perhaps these are all the possibilities.

But let me check if there are other quadratics. For example, suppose that for both roots, $c^2 - 2$ is the other root. So, for example, if we have roots $r$ and $s$, such that $r^2 - 2 = s$ and $s^2 - 2 = r$. Which is Subcase 2b. Which gives us the quadratic $x^2 + x - 1$. Also, in Subcase 2c, we have quadratics where one root maps to itself and the other maps to the first root. So, for example, if one root is 2 (which maps to itself), and the other root is -2 (which maps to 2). Similarly, if one root is -1 (which maps to itself), and the other root is 1 (which maps to -1). Therefore, these quadratics are $x^2 - 4$ and $x^2 - 1$.

Additionally, in Subcase 2a, we have the quadratic with roots 2 and -1, which both map to themselves. Therefore, that quadratic is $x^2 - x - 2$.

So, so far, we have:

From Case 1: 2 quadratics.

From Case 2a: 1 quadratic.

From Case 2b: 1 quadratic.

From Case 2c: 2 quadratics.

Wait, but hold on. Let me check if these quadratics are distinct. Let me list them:

Case 1:

1. $x^2 - 4x + 4$ (roots 2, 2)

2. $x^2 + 2x + 1$ (roots -1, -1)

Case 2a:

3. $x^2 - x - 2$ (roots 2, -1)

Case 2b:

4. $x^2 + x - 1$ (roots $[-1 \pm \sqrt{5}]/2$)

Case 2c:

5. $x^2 - 4$ (roots 2, -2)

6. $x^2 - 1$ (roots 1, -1)

So, that's 6 quadratics. Wait, but let me check if there are more possibilities. For example, in Subcase 2c, when we considered $s = 2$ and $s = -1$, we found quadratics with roots 2 and -2, and 1 and -1. Are there other possibilities?

Wait, for example, if we take $s = 2$, then $r^2 - 2 = 2$ gives $r = \pm 2$. If $r = 2$, then we have repeated roots. If $r = -2$, then we have roots -2 and 2. Similarly, for $s = -1$, $r^2 - 2 = -1$ gives $r = \pm 1$. If $r = 1$, then roots are 1 and -1. If $r = -1$, then repeated roots. Therefore, only two new quadratics here: $x^2 - 4$ and $x^2 - 1$.

Is there another case where, for example, both roots map to each other, but not covered in Subcase 2b? For example, if we have roots $r$ and $s$ such that $r^2 - 2 = s$ and $s^2 - 2 = r$, which is exactly Subcase 2b. So that's already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That would be Case 2a, which is already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to another root? Which is Subcase 2c.

Therefore, perhaps these are all the possibilities. Let me check if there are any other quadratics.

Wait, let me check if the quadratic $x^2 - 1$ is already considered. Yes, in Case 2c. Similarly, $x^2 - 4$ is in Case 2c. The quadratic $x^2 - x - 2$ is in Case 2a. The quadratic $x^2 + x - 1$ is in Case 2b. The two quadratics from Case 1 are already there.

But let me check if there are other quadratics where, for example, both roots map to each other in some other way. For example, suppose that $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? Wait, that's exactly Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, could there be quadratics where one root maps to another root, and that root maps to another root? For example, if we have three roots? But since it's a quadratic, there are only two roots. So, if $r^2 - 2 = s$, and $s^2 - 2 = r$, then applying the transformation twice, we get $r^4 - 4r^2 + 4 - 2 = r^4 - 4r^2 + 2 = r$. Which is the same quartic equation as before, leading to the roots we already considered. Therefore, no new quadratics there.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That's Case 2a, which gives roots 2 and -1. So that's already considered.

Therefore, so far, we have 6 quadratics. Wait, but let me check if there are any overlaps or duplicates. For example, is $x^2 - 1$ the same as any other quadratic? No. Similarly, $x^2 - 4$ is unique. The quadratic $x^2 - x - 2$ is unique. The quadratic $x^2 + x - 1$ is unique. The two quadratics from Case 1 are unique. Therefore, total of 6 quadratics.

But let me check if there are more quadratics. For example, suppose that we have a quadratic with roots $r$ and $s$, where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? Wait, that's exactly Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, suppose that one root maps to another root, and that root maps to another root, but since there are only two roots, this would require that applying the transformation twice brings you back. For example, if $r^2 - 2 = s$ and $s^2 - 2 = r$, which is the same as before. So, that's already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to another root, but not covered in Subcase 2c? For example, suppose that one root is 2 (which maps to itself) and the other root is something else. Wait, but if one root is 2, then the other root must satisfy that $c^2 - 2$ is a root. Since 2 is already a root, then the other root must be either 2 or something else. If the other root is 2, then it's a repeated root. If the other root is something else, say $s$, then $s^2 - 2$ must be a root. Since the roots are 2 and $s$, then $s^2 - 2$ must be either 2 or $s$. If $s^2 - 2 = 2$, then $s^2 = 4$, so $s = \pm 2$. If $s = 2$, then repeated root. If $s = -2$, then we have roots 2 and -2, which is the quadratic $x^2 - 4$, which is already considered. If $s^2 - 2 = s$, then $s^2 - s - 2 = 0$, which gives $s = 2$ or $s = -1$. If $s = 2$, repeated root. If $s = -1$, then roots are 2 and -1, which is quadratic $x^2 - x - 2$, already considered. Therefore, no new quadratics here.

Similarly, if one root is -1, then the other root must satisfy $s^2 - 2$ is a root. If the other root is -1, then repeated root. If the other root is something else, say $s$, then $s^2 - 2$ must be either -1 or $s$. If $s^2 - 2 = -1$, then $s^2 = 1$, so $s = \pm 1$. If $s = 1$, then roots are -1 and 1, quadratic $x^2 - 1$, already considered. If $s = -1$, repeated root. If $s^2 - 2 = s$, then $s^2 - s - 2 = 0$, which gives $s = 2$ or $s = -1$. If $s = 2$, then roots are -1 and 2, quadratic $x^2 - x - 2$, already considered. If $s = -1$, repeated root. Therefore, again, no new quadratics.

Therefore, all possibilities seem to be covered. Therefore, total of 6 quadratics. Wait, but let me check again:

Case 1: 2 quadratics.

Case 2a: 1 quadratic.

Case 2b: 1 quadratic.

Case 2c: 2 quadratics.

Total: 2 + 1 + 1 + 2 = 6.

But let me check if there are any other quadratics. For example, suppose that we have a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? Wait, that's exactly Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, could there be quadratics where both roots map to each other in some other way? For example, if $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? But that's the same as Subcase 2b. So, no.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That's Case 2a, which gives roots 2 and -1. So that's already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to another root, but not covered in Case 2c? For example, if one root is 2 (maps to itself) and the other root is something else. But we already considered that in Case 2c. Similarly for -1.

Therefore, I think we have all the quadratics. Therefore, total of 6 quadratics. However, let me check each quadratic to ensure they satisfy the condition.

First, Case 1:

1. $x^2 - 4x + 4$: roots 2, 2. For root 2: 2^2 - 2 = 2, which is a root. Good.

2. $x^2 + 2x + 1$: roots -1, -1. For root -1: (-1)^2 - 2 = -1, which is a root. Good.

Case 2a:

3. $x^2 - x - 2$: roots 2, -1. For root 2: 2^2 - 2 = 2, which is a root. For root -1: (-1)^2 - 2 = -1, which is a root. Good.

Case 2b:

4. $x^2 + x - 1$: roots $[-1 \pm \sqrt{5}]/2$. Let me denote them as $r_1$ and $r_2$. For $r_1$, $r_1^2 - 2 = r_2$, which is a root. Similarly for $r_2$, $r_2^2 - 2 = r_1$. Good.

Case 2c:

5. $x^2 - 4$: roots 2, -2. For root 2: 2^2 - 2 = 2, which is a root. For root -2: (-2)^2 - 2 = 2, which is a root. Good.

6. $x^2 - 1$: roots 1, -1. For root 1: 1^2 - 2 = -1, which is a root. For root -1: (-1)^2 - 2 = -1, which is a root. Good.

Therefore, all six quadratics satisfy the condition. Therefore, the total number is 6. However, let me check if there are any other quadratics that I might have missed.

Wait, let me think again. For example, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots. Wait, that's exactly the quadratic $x^2 + x - 1$, which we already considered. So, no new quadratics there.

Alternatively, could there be quadratics where one root maps to another root, and that root maps to another root, but not forming a cycle? But since there are only two roots, if $r^2 - 2 = s$ and $s^2 - 2 = r$, then applying the transformation twice gives $r^4 - 4r^2 + 2 = r$, which is the quartic equation we solved earlier. Therefore, the roots of that quartic are already considered. Therefore, no new quadratics.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That's Case 2a, which gives roots 2 and -1. So that's already considered.

Therefore, I think we have all the quadratics. Therefore, the answer is 6. However, let me check once more for possible duplicates or errors.

Wait, but let me check if the quadratic $x^2 + x - 1$ is distinct from the others. Yes, because its roots are different from the others. Similarly, $x^2 - 4$ and $x^2 - 1$ are distinct. The quadratics in Case 1 are distinct from the others. The quadratic $x^2 - x - 2$ is distinct. Therefore, total of 6 quadratics.

But wait, hold on. Let me check if there are quadratics with different coefficients but same roots. For example, if two different quadratics have the same roots, but since we are considering monic quadratics (since the problem says "quadratic equations of the form $x^2 + ax + b = 0$"), then each set of roots corresponds to exactly one quadratic. Therefore, each distinct set of roots gives a unique quadratic. Therefore, since we have 6 distinct quadratics, the answer is 6.

But let me check once more if there are other quadratics. For example, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots. Wait, but that's exactly the quadratic $x^2 + x - 1$, which we already considered. Therefore, no.

Alternatively, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with $r$ and $s$ being different from the ones in $x^2 + x - 1$. But solving $r^2 - 2 = s$ and $s^2 - 2 = r$ leads to the quartic equation which factors into (r + 1)(r - 2)(r^2 + r - 1) = 0. Therefore, the roots are -1, 2, and the roots of $r^2 + r - 1 = 0$. Therefore, the only possible pairs of roots are:

- (-1, -1) [repeated root]

- (2, 2) [repeated root]

- (-1, 2) [distinct roots]

- (roots of $r^2 + r - 1 = 0$) [distinct roots]

- (-1, -2) [from Subcase 2c]

- (1, -1) [from Subcase 2c]

Wait, but how does (-1, -2) come into play? Let me think. If we have roots -1 and -2, then for root -1: (-1)^2 - 2 = -1, which is a root. For root -2: (-2)^2 - 2 = 2, which is not a root. Wait, hold on! Wait, earlier I thought that quadratic $x^2 - 4$ has roots 2 and -2. Let me check again. If the quadratic is $x^2 - 4$, then the roots are 2 and -2. For root 2: 2^2 - 2 = 2, which is a root. For root -2: (-2)^2 - 2 = 4 - 2 = 2, which is a root. Therefore, yes, that works. But if we have roots -1 and -2, then for root -2, $c^2 - 2 = 2$, which is not a root unless 2 is a root. But if the quadratic is $x^2 - 4$, then 2 is a root. Therefore, the roots are 2 and -2. Therefore, the quadratic $x^2 - 4$ is valid. However, if we have roots -1 and -2, then the quadratic would be $x^2 - (-1 + (-2))x + (-1)(-2) = x^2 + 3x + 2$. Let me check if this quadratic satisfies the condition. For root -1: (-1)^2 - 2 = -1, which is a root. For root -2: (-2)^2 - 2 = 2, which is not a root of $x^2 + 3x + 2$. Therefore, this quadratic does not satisfy the condition. Therefore, my earlier reasoning was wrong. Wait, how did I get the quadratic $x^2 - 4$?

Earlier, in Subcase 2c, when I considered $s = 2$ and $r^2 - 2 = 2$, leading to $r = \pm 2$. If $r = -2$, then the roots are -2 and 2, which is the quadratic $x^2 - 4$. Therefore, that quadratic is valid. However, if I have roots -1 and -2, that quadratic is not valid. Therefore, my mistake earlier was thinking that if $s = -1$, then $r^2 - 2 = -1$ gives $r = \pm 1$, leading to roots 1 and -1. Therefore, quadratic $x^2 - 1$. Therefore, the quadratic $x^2 - 1$ has roots 1 and -1. For root 1: 1^2 - 2 = -1, which is a root. For root -1: (-1)^2 - 2 = -1, which is a root. Therefore, that quadratic is valid. However, if I have roots -1 and -2, that quadratic is invalid. Therefore, my earlier reasoning was correct in that only quadratics with roots 2 and -2, and 1 and -1 are valid. Therefore, the quadratics $x^2 - 4$ and $x^2 - 1$ are valid. Therefore, my previous conclusion of 6 quadratics is correct.

Therefore, the total number of such quadratics is 6. However, let me check once more if there are any other quadratics.

Wait, let me check if the quadratic $x^2 + x - 1$ is the only one from Subcase 2b. Yes, because the roots are from the equation $x^2 + x - 1 = 0$, which is unique. Therefore, that's one quadratic.

Therefore, total of 6 quadratics. Therefore, the answer is 6. However, let me check once more for possible mistakes.

Wait, but let me check if the quadratic $x^2 + x - 1$ is the only one from Subcase 2b. Yes, because the quartic equation had roots -1, 2, and the roots of $x^2 + x - 1 = 0$. Therefore, the only pair of roots that are not repeated and not involving -1 or 2 is the roots of $x^2 + x - 1 = 0$. Therefore, that's one quadratic.

Therefore, total of 6 quadratics. Therefore, the answer is 6. However, let me check if the problem says "quadratic equations of the form $x^2 + ax + b = 0$", so they are monic quadratics. Therefore, each distinct pair of roots gives a unique quadratic. Therefore, 6 quadratics. Therefore, the answer is 6.

But wait, let me check once more for possible duplicates. For example, is $x^2 - x - 2$ the same as any other quadratic? No. Similarly, $x^2 + x - 1$ is unique. The quadratics in Case 1 are unique. The quadratics in Case 2c are unique. Therefore, total of 6.

But let me check if there are other quadratics where, for example, both roots map to each other in some other way. For example, suppose that $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots. Wait, but that's exactly the quadratic $x^2 + x - 1$, which we already considered. Therefore, no.

Alternatively, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with $r$ and $s$ being different from the ones in $x^2 + x - 1$. But solving $r^2 - 2 = s$ and $s^2 - 2 = r$ leads to the quartic equation, which only has those roots. Therefore, no.

Therefore, I think the answer is 6. However, I need to check if the problem counts quadratics with the same roots but different coefficients. But since the problem says "quadratic equations of the form $x^2 + ax + b = 0$", each distinct pair of roots gives a unique quadratic. Therefore, 6 quadratics.

But wait, let me check once more. Let me list all six quadratics:

1. $x^2 - 4x + 4$ (a = -4, b = 4)

2. $x^2 + 2x + 1$ (a = 2, b = 1)

3. $x^2 - x - 2$ (a = -1, b = -2)

4. $x^2 + x - 1$ (a = 1, b = -1)

5. $x^2 - 4$ (a = 0, b = -4)

6. $x^2 - 1$ (a = 0, b = -1)

Yes, all these have different coefficients (a, b). Therefore, they are distinct equations. Therefore, the total number is 6.

But wait, hold on. The quadratic $x^2 - 4$ has a = 0, b = -4. The quadratic $x^2 - 1$ has a = 0, b = -1. So they are different. Similarly, the others have different a and b. Therefore, 6 distinct quadratics.

Therefore, the answer is 6. However, I need to check if there are any other quadratics that I might have missed. For example, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with $r$ and $s$ being complex numbers? Wait, but the problem doesn't specify that the roots have to be real. However, the problem says "quadratic equations of the form $x^2 + ax + b = 0$", and doesn't specify that the coefficients a and b are real numbers. Wait, but usually, in such problems, unless specified otherwise, we assume that a and b are real numbers. However, even if they are complex, the problem might still have solutions. However, the condition is that whenever c is a root, c^2 - 2 is also a root. If the roots are complex, then c^2 - 2 might not be a root unless specific conditions are met. However, since the problem doesn't specify, but given that the answer is likely to be an integer, and given that the previous cases have real roots, maybe the problem assumes real coefficients and real roots? Or maybe complex roots are allowed?

Wait, but if complex roots are allowed, then there might be more quadratics. However, since the problem is likely from a competition or textbook problem, it's more probable that they are considering real coefficients and real roots. However, let me check.

Suppose that we allow complex roots. Then, for example, the quadratic $x^2 + x - 1 = 0$ has complex roots? No, wait, discriminant is $1 + 4 = 5$, which is positive, so real roots. The quadratic $x^2 + x - 1$ has real roots. The quadratic $x^2 - x - 2$ has real roots. The quadratics in Case 1 have real roots. The quadratics in Case 2c have real roots. Therefore, all quadratics considered so far have real roots. Therefore, even if complex roots are allowed, there are no additional quadratics because the other roots would be complex and would not satisfy the condition unless they are closed under the transformation $c \mapsto c^2 - 2$. However, since the problem didn't specify, but given that all the quadratics we found have real roots, and the answer is 6, I think that is the intended answer.

Therefore, I think the number of such quadratic equations is 6. Therefore, the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To solve the problem, we are asked to find the number of **monic quadratic equations** of the form:

$$
x^2 + ax + b = 0
$$

such that **whenever $ c $ is a root**, then $ c^2 - 2 $ is also a root.

---

### Step 1: Understanding the Condition

Let the roots of the quadratic be $ r_1 $ and $ r_2 $. The condition implies that for each root $ c $, the value $ c^2 - 2 $ must also be a root. Since the quadratic has only two roots, this condition must be satisfied for both roots.

This leads to the following possibilities:

- If $ r_1 = r_2 $, then the quadratic has a **repeated root**. In this case, the condition implies that $ r^2 - 2 = r $, which gives a quadratic equation in $ r $:
  $$
  r^2 - r - 2 = 0 \Rightarrow (r - 2)(r + 1) = 0
  $$
  So the repeated roots are $ r = 2 $ or $ r = -1 $, leading to two quadratics:
  - $ (x - 2)^2 = x^2 - 4x + 4 $
  - $ (x + 1)^2 = x^2 + 2x + 1 $

- If $ r_1 \ne r_2 $, then the roots must satisfy the condition that $ r_1^2 - 2 = r_2 $ and $ r_2^2 - 2 = r_1 $. This leads to the system:
  $$
  r_1^2 - 2 = r_2 \quad \text{and} \quad r_2^2 - 2 = r_1
  $$
  Substituting $ r_2 = r_1^2 - 2 $ into the second equation gives:
  $$
  (r_1^2 - 2)^2 - 2 = r_1 \Rightarrow r_1^4 - 4r_1^2 - r_1 + 2 = 0
  $$
  Factoring this quartic yields:
  $$
  (r + 1)(r - 2)(r^2 + r - 1) = 0
  $$
  The roots are:
  - $ r = -1 $, $ r = 2 $, and the roots of $ r^2 + r - 1 = 0 $, which are:
    $$
    r = \frac{-1 \pm \sqrt{5}}{2}
    $$

  From this, we can form the following distinct quadratics:

  - Roots $ r = 2 $ and $ r = -1 $: $ x^2 - x - 2 $
  - Roots $ r = \frac{-1 + \sqrt{5}}{2} $ and $ r = \frac{-1 - \sqrt{5}}{2} $: $ x^2 + x - 1 $

  Additionally, we consider the case where one root maps to itself and the other maps to the first root. For example:

  - If $ r = 2 $, then $ r^2 - 2 = 2 $, so the other root must be $ -2 $, leading to the quadratic $ x^2 - 4 $
  - If $ r = -1 $, then $ r^2 - 2 = -1 $, so the other root must be $ 1 $, leading to the quadratic $ x^2 - 1 $

---

### Step 2: List All Valid Quadratics

From the above, we have the following **six distinct monic quadratic equations** that satisfy the condition:

1. $ x^2 - 4x + 4 $ (repeated root 2)
2. $ x^2 + 2x + 1 $ (repeated root -1)
3. $ x^2 - x - 2 $ (roots 2 and -1)
4. $ x^2 + x - 1 $ (roots $ \frac{-1 \pm \sqrt{5}}{2} $)
5. $ x^2 - 4 $ (roots 2 and -2)
6. $ x^2 - 1 $ (roots 1 and -1)

Each of these quadratics has been verified to satisfy the condition that if $ c $ is a root, then $ c^2 - 2 $ is also a root.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the number of quadratic equations of the form $x^2 + ax + b = 0$ such that whenever $c$ is a root, $c^2 - 2$ is also a root. Hmm, let me think about how to approach this.

First, let me recall that a quadratic equation has two roots (counting multiplicity). Let me denote the roots as $r_1$ and $r_2$. According to Vieta's formulas, we have:

$r_1 + r_2 = -a$ and $r_1 r_2 = b$.

Now, the condition given is that if $c$ is a root, then $c^2 - 2$ is also a root. So, for each root $c$, $c^2 - 2$ must be another root. Since there are only two roots, this condition must hold for both roots. Let me consider the possibilities.

Case 1: The quadratic has a repeated root. That is, $r_1 = r_2$. Then, if $c = r_1$, then $c^2 - 2$ must also be a root. But since there's only one root, $c^2 - 2$ must equal $c$. So, solving $c^2 - 2 = c$, which gives $c^2 - c - 2 = 0$. Solving this quadratic equation:

$c = [1 \pm \sqrt{1 + 8}]/2 = [1 \pm 3]/2$, so $c = 2$ or $c = -1$.

Therefore, if the quadratic has a repeated root, it must be either 2 or -1. Let me check if these satisfy the condition.

If the root is 2, then $c^2 - 2 = 4 - 2 = 2$, which is the same root. So that works. Similarly, if the root is -1, then $c^2 - 2 = 1 - 2 = -1$, which is also the same root. So both 2 and -1 are valid repeated roots. Therefore, there are two quadratics in this case: $(x - 2)^2 = x^2 - 4x + 4$ and $(x + 1)^2 = x^2 + 2x + 1$. So that's two equations.

Case 2: The quadratic has two distinct roots. Let me denote them as $r$ and $s$. Then, according to the condition, if $r$ is a root, then $r^2 - 2$ must be a root. Similarly, if $s$ is a root, then $s^2 - 2$ must be a root. Since there are only two roots, this implies that $r^2 - 2$ must be either $r$ or $s$, and similarly for $s^2 - 2$.

So, let me consider the possibilities for each root:

For root $r$, $r^2 - 2$ is either $r$ or $s$.

Similarly, for root $s$, $s^2 - 2$ is either $r$ or $s$.

Let me consider different subcases.

Subcase 2a: Suppose $r^2 - 2 = r$ and $s^2 - 2 = s$. Then, this would imply that both roots satisfy $x^2 - x - 2 = 0$, which factors as $(x - 2)(x + 1) = 0$. Therefore, the roots would be 2 and -1. But then, this is the case where the quadratic has roots 2 and -1. However, in this case, the quadratic would be $(x - 2)(x + 1) = x^2 - x - 2$. But wait, this is a quadratic with distinct roots. However, we need to check if this satisfies the original condition. Let me check:

If $c = 2$, then $c^2 - 2 = 4 - 2 = 2$, which is a root. If $c = -1$, then $c^2 - 2 = 1 - 2 = -1$, which is also a root. So this quadratic does satisfy the condition. However, this quadratic is different from the ones in Case 1. Wait, but in Case 1, we had quadratics with repeated roots. So this is a separate case. Therefore, this quadratic is valid. However, note that this quadratic is different from the ones in Case 1. So this is another quadratic. But wait, but in this case, the roots are 2 and -1, so the quadratic is $x^2 - x - 2$. However, in Case 1, we had quadratics with repeated roots 2 and -1. So this is a different quadratic. Therefore, this is another case. However, let me check if there are other possibilities.

Subcase 2b: Suppose that for one root, $r^2 - 2 = s$, and for the other root, $s^2 - 2 = r$. So, this would mean that applying the transformation $c \mapsto c^2 - 2$ swaps the roots. So, we have:

$r^2 - 2 = s$ and $s^2 - 2 = r$.

So, we can set up these equations:

From the first equation: $s = r^2 - 2$.

Substitute into the second equation: $(r^2 - 2)^2 - 2 = r$.

Let me expand that:

$(r^4 - 4r^2 + 4) - 2 = r$

Simplify:

$r^4 - 4r^2 + 2 = r$

Bring all terms to one side:

$r^4 - 4r^2 - r + 2 = 0$

Hmm, quartic equation. Let me try to factor this. Maybe there are rational roots. Let me apply Rational Root Theorem. Possible rational roots are ±1, ±2.

Testing r = 1: 1 - 4 - 1 + 2 = -2 ≠ 0.

Testing r = -1: 1 - 4 + 1 + 2 = 0. Oh, r = -1 is a root.

So, we can factor (r + 1) out of the quartic.

Let me perform polynomial division or use synthetic division.

Divide $r^4 - 4r^2 - r + 2$ by (r + 1).

Using synthetic division for root -1:

Coefficients: 1 (r^4), 0 (r^3), -4 (r^2), -1 (r), 2 (constant).

Bring down the 1.

Multiply by -1: -1. Add to next coefficient: 0 + (-1) = -1.

Multiply by -1: 1. Add to next coefficient: -4 + 1 = -3.

Multiply by -1: 3. Add to next coefficient: -1 + 3 = 2.

Multiply by -1: -2. Add to last coefficient: 2 + (-2) = 0. Perfect.

So the quartic factors as (r + 1)(r^3 - r^2 - 3r + 2).

Now, let's factor the cubic: $r^3 - r^2 - 3r + 2$.

Again, try rational roots: ±1, ±2.

Testing r = 1: 1 - 1 - 3 + 2 = -1 ≠ 0.

Testing r = 2: 8 - 4 - 6 + 2 = 0. Yes, r = 2 is a root.

Divide the cubic by (r - 2):

Using synthetic division:

Coefficients: 1 (r^3), -1 (r^2), -3 (r), 2 (constant).

Root at 2:

Bring down 1.

Multiply by 2: 2. Add to -1: 1.

Multiply by 2: 2. Add to -3: -1.

Multiply by 2: -2. Add to 2: 0. Perfect.

So the cubic factors as (r - 2)(r^2 + r - 1).

Therefore, the quartic factors as (r + 1)(r - 2)(r^2 + r - 1).

Therefore, the roots of the quartic equation are r = -1, r = 2, and roots of $r^2 + r - 1 = 0$, which are $r = [-1 \pm \sqrt{1 + 4}]/2 = [-1 \pm \sqrt{5}]/2$.

So, going back to our original equations, we had:

s = r^2 - 2, and s^2 - 2 = r.

So, for each root r of the quartic, we can find s. However, we need to check if these lead to valid quadratics.

First, let's consider r = -1. Then s = (-1)^2 - 2 = 1 - 2 = -1. Therefore, s = -1. So, this gives us the case where both roots are -1, which is the repeated root case we considered earlier. So this is already covered in Case 1.

Similarly, for r = 2: s = 2^2 - 2 = 4 - 2 = 2. Again, this gives s = 2, which is the repeated root case. So again, covered in Case 1.

Now, for the roots of $r^2 + r - 1 = 0$, which are $r = [-1 + \sqrt{5}]/2$ and $r = [-1 - \sqrt{5}]/2$. Let me denote them as $r_1$ and $r_2$.

So, for each of these roots, we can compute s = r^2 - 2. Let me compute s for $r_1$:

First, $r_1 = [-1 + \sqrt{5}]/2$. Let me compute $r_1^2$:

$r_1^2 = [(-1 + \sqrt{5})^2]/4 = [1 - 2\sqrt{5} + 5]/4 = [6 - 2\sqrt{5}]/4 = [3 - \sqrt{5}]/2$.

Then, $s = r_1^2 - 2 = [3 - \sqrt{5}]/2 - 2 = [3 - \sqrt{5} - 4]/2 = [-1 - \sqrt{5}]/2 = r_2$.

Similarly, for $r_2 = [-1 - \sqrt{5}]/2$, compute $r_2^2$:

$r_2^2 = [(-1 - \sqrt{5})^2]/4 = [1 + 2\sqrt{5} + 5]/4 = [6 + 2\sqrt{5}]/4 = [3 + \sqrt{5}]/2$.

Then, $s = r_2^2 - 2 = [3 + \sqrt{5}]/2 - 2 = [3 + \sqrt{5} - 4]/2 = [-1 + \sqrt{5}]/2 = r_1$.

Therefore, for these roots, we have that $s = r_2$ when $r = r_1$, and $s = r_1$ when $r = r_2$. Therefore, this gives us a pair of roots $r_1$ and $r_2$, which are distinct. Therefore, this corresponds to a quadratic equation with roots $r_1$ and $r_2$. Let me write down the quadratic equation.

Since the roots are $r_1$ and $r_2$, the quadratic is $x^2 - (r_1 + r_2)x + r_1 r_2$. Let me compute $r_1 + r_2$ and $r_1 r_2$.

From the quadratic equation $r^2 + r - 1 = 0$, we know that $r_1 + r_2 = -1$ and $r_1 r_2 = -1$. Therefore, the quadratic equation is $x^2 - (-1)x + (-1) = x^2 + x - 1$. Therefore, this quadratic is $x^2 + x - 1$. Let me check if this satisfies the original condition.

Take root $r_1 = [-1 + \sqrt{5}]/2$. Compute $c^2 - 2$:

As above, $c^2 - 2 = r_2$, which is the other root. Similarly for $r_2$, $c^2 - 2 = r_1$. Therefore, this quadratic satisfies the condition. Therefore, this is another valid quadratic equation.

So, in Subcase 2b, we have another quadratic equation $x^2 + x - 1$.

Wait, but hold on. Earlier, in Subcase 2a, we had the quadratic $x^2 - x - 2$ with roots 2 and -1. Let me check if that quadratic also satisfies the condition. Let me check if for each root, applying $c^2 - 2$ gives another root.

For root 2: $2^2 - 2 = 4 - 2 = 2$, which is the same root. For root -1: $(-1)^2 - 2 = 1 - 2 = -1$, same root. Therefore, this quadratic also satisfies the condition. However, in Subcase 2a, we considered the case where both roots satisfy $c^2 - 2 = c$, which led us to the roots 2 and -1. However, in this case, the quadratic is $x^2 - x - 2$, which is different from the quadratics in Case 1 (which had repeated roots). Therefore, this is another quadratic. So, in total, we have:

Case 1: two quadratics (repeated roots 2 and -1).

Case 2a: one quadratic (roots 2 and -1).

Case 2b: one quadratic (roots $r_1$ and $r_2$ from the equation $x^2 + x - 1 = 0$).

Wait, but hold on. Wait, in Case 2a, we considered the case where both roots satisfy $c^2 - 2 = c$, which gives roots 2 and -1. However, in Case 2b, we have another quadratic with roots $r_1$ and $r_2$ which are not 2 or -1. So, that's a separate quadratic. However, is there another case?

Wait, let me check if there are other possibilities. For example, in Subcase 2c: Suppose that for one root, $r^2 - 2 = r$, and for the other root, $s^2 - 2 = s$. Wait, that's exactly Subcase 2a. So that's already considered.

Alternatively, maybe there's another case where one root maps to itself, and the other maps to the first root? Wait, but if $r^2 - 2 = r$, then that's covered in Subcase 2a. If $s^2 - 2 = r$, then that's covered in Subcase 2b. So, perhaps those are the only possibilities.

Wait, but let me think again. Suppose that for one root, $r^2 - 2 = s$, and for the other root, $s^2 - 2 = s$. Then, that would be another case. Let me check.

Subcase 2c: Suppose $r^2 - 2 = s$ and $s^2 - 2 = s$. Then, from the second equation, $s^2 - s - 2 = 0$, which factors as $(s - 2)(s + 1) = 0$, so $s = 2$ or $s = -1$. Then, from the first equation, $r^2 - 2 = s$. Therefore, if $s = 2$, then $r^2 - 2 = 2$ implies $r^2 = 4$, so $r = 2$ or $r = -2$. Similarly, if $s = -1$, then $r^2 - 2 = -1$ implies $r^2 = 1$, so $r = 1$ or $r = -1$.

So, let's check these possibilities.

First, if $s = 2$, then $r$ can be 2 or -2. If $r = 2$, then the roots are 2 and 2, which is the repeated root case (already considered in Case 1). If $r = -2$, then the roots are -2 and 2. Let me check if this quadratic satisfies the condition. Let me check for each root:

For root 2: $2^2 - 2 = 2$, which is a root. For root -2: $(-2)^2 - 2 = 4 - 2 = 2$, which is a root. Therefore, this quadratic $x^2 - (2 + (-2))x + (2)(-2) = x^2 - 0x - 4 = x^2 - 4$. Wait, but hold on, if the roots are 2 and -2, then Vieta's formula gives sum = 0, product = -4. Therefore, the quadratic is $x^2 - 0x - 4 = x^2 - 4$. Let me check if this quadratic satisfies the condition. For root 2: 2^2 - 2 = 2, which is a root. For root -2: (-2)^2 - 2 = 4 - 2 = 2, which is a root. Therefore, yes, this quadratic satisfies the condition. Therefore, this is another quadratic equation. Similarly, if $s = -1$, then $r^2 - 2 = -1$ implies $r^2 = 1$, so $r = 1$ or $r = -1$. If $r = 1$, then the roots are 1 and -1. Let me check if this quadratic satisfies the condition. The quadratic would be $x^2 - (1 + (-1))x + (1)(-1) = x^2 - 0x -1 = x^2 - 1$. Let me check the roots: 1 and -1. For root 1: 1^2 - 2 = -1, which is a root. For root -1: (-1)^2 - 2 = 1 - 2 = -1, which is a root. Therefore, this quadratic also satisfies the condition. Similarly, if $r = -1$, then the roots are -1 and -1, which is the repeated root case already considered. Therefore, in this Subcase 2c, we have two new quadratics: $x^2 - 4$ and $x^2 - 1$.

Wait, so this seems like another case. Therefore, I need to check if these quadratics were already considered in previous cases. In Case 1, we had quadratics with repeated roots 2 and -1. Here, we have quadratics with distinct roots 2 and -2, and 1 and -1. Therefore, these are new quadratics. Therefore, this suggests that there are more quadratics to consider.

So, perhaps my initial analysis was incomplete. Let me try to organize all these cases.

So, let me recap:

Case 1: Repeated roots. We had quadratics with roots 2 and 2, and -1 and -1. So two quadratics.

Case 2: Distinct roots. Let me consider different possibilities:

Subcase 2a: Both roots satisfy $c^2 - 2 = c$. This gives roots 2 and -1, leading to quadratic $x^2 - x - 2$.

Subcase 2b: The roots are such that $r^2 - 2 = s$ and $s^2 - 2 = r$. This gives roots from $x^2 + x - 1 = 0$, leading to quadratic $x^2 + x - 1$.

Subcase 2c: One root satisfies $c^2 - 2 = c$ (i.e., 2 or -1) and the other root maps to the first root. For example, if one root is 2, then the other root is -2 (as in $x^2 - 4$), or if one root is -1, the other root is 1 (as in $x^2 - 1$). Wait, but in Subcase 2c, we considered when one root maps to itself (i.e., satisfies $c^2 - 2 = c$) and the other maps to the first root. Therefore, these quadratics $x^2 - 4$ and $x^2 - 1$ are valid.

But wait, are there more possibilities? For example, if one root maps to the other root, and the other root maps to itself? Wait, that's what Subcase 2c is. So, in addition to that, maybe there are other quadratics where neither root maps to itself, but they map to each other? Wait, that's Subcase 2b. So, perhaps these are all the possibilities.

But let me check if there are other quadratics. For example, suppose that for both roots, $c^2 - 2$ is the other root. So, for example, if we have roots $r$ and $s$, such that $r^2 - 2 = s$ and $s^2 - 2 = r$. Which is Subcase 2b. Which gives us the quadratic $x^2 + x - 1$. Also, in Subcase 2c, we have quadratics where one root maps to itself and the other maps to the first root. So, for example, if one root is 2 (which maps to itself), and the other root is -2 (which maps to 2). Similarly, if one root is -1 (which maps to itself), and the other root is 1 (which maps to -1). Therefore, these quadratics are $x^2 - 4$ and $x^2 - 1$.

Additionally, in Subcase 2a, we have the quadratic with roots 2 and -1, which both map to themselves. Therefore, that quadratic is $x^2 - x - 2$.

So, so far, we have:

From Case 1: 2 quadratics.

From Case 2a: 1 quadratic.

From Case 2b: 1 quadratic.

From Case 2c: 2 quadratics.

Wait, but hold on. Let me check if these quadratics are distinct. Let me list them:

Case 1:

1. $x^2 - 4x + 4$ (roots 2, 2)

2. $x^2 + 2x + 1$ (roots -1, -1)

Case 2a:

3. $x^2 - x - 2$ (roots 2, -1)

Case 2b:

4. $x^2 + x - 1$ (roots $[-1 \pm \sqrt{5}]/2$)

Case 2c:

5. $x^2 - 4$ (roots 2, -2)

6. $x^2 - 1$ (roots 1, -1)

So, that's 6 quadratics. Wait, but let me check if there are more possibilities. For example, in Subcase 2c, when we considered $s = 2$ and $s = -1$, we found quadratics with roots 2 and -2, and 1 and -1. Are there other possibilities?

Wait, for example, if we take $s = 2$, then $r^2 - 2 = 2$ gives $r = \pm 2$. If $r = 2$, then we have repeated roots. If $r = -2$, then we have roots -2 and 2. Similarly, for $s = -1$, $r^2 - 2 = -1$ gives $r = \pm 1$. If $r = 1$, then roots are 1 and -1. If $r = -1$, then repeated roots. Therefore, only two new quadratics here: $x^2 - 4$ and $x^2 - 1$.

Is there another case where, for example, both roots map to each other, but not covered in Subcase 2b? For example, if we have roots $r$ and $s$ such that $r^2 - 2 = s$ and $s^2 - 2 = r$, which is exactly Subcase 2b. So that's already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That would be Case 2a, which is already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to another root? Which is Subcase 2c.

Therefore, perhaps these are all the possibilities. Let me check if there are any other quadratics.

Wait, let me check if the quadratic $x^2 - 1$ is already considered. Yes, in Case 2c. Similarly, $x^2 - 4$ is in Case 2c. The quadratic $x^2 - x - 2$ is in Case 2a. The quadratic $x^2 + x - 1$ is in Case 2b. The two quadratics from Case 1 are already there.

But let me check if there are other quadratics where, for example, both roots map to each other in some other way. For example, suppose that $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? Wait, that's exactly Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, could there be quadratics where one root maps to another root, and that root maps to another root? For example, if we have three roots? But since it's a quadratic, there are only two roots. So, if $r^2 - 2 = s$, and $s^2 - 2 = r$, then applying the transformation twice, we get $r^4 - 4r^2 + 4 - 2 = r^4 - 4r^2 + 2 = r$. Which is the same quartic equation as before, leading to the roots we already considered. Therefore, no new quadratics there.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That's Case 2a, which gives roots 2 and -1. So that's already considered.

Therefore, so far, we have 6 quadratics. Wait, but let me check if there are any overlaps or duplicates. For example, is $x^2 - 1$ the same as any other quadratic? No. Similarly, $x^2 - 4$ is unique. The quadratic $x^2 - x - 2$ is unique. The quadratic $x^2 + x - 1$ is unique. The two quadratics from Case 1 are unique. Therefore, total of 6 quadratics.

But let me check if there are more quadratics. For example, suppose that we have a quadratic with roots $r$ and $s$, where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? Wait, that's exactly Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, suppose that one root maps to another root, and that root maps to another root, but since there are only two roots, this would require that applying the transformation twice brings you back. For example, if $r^2 - 2 = s$ and $s^2 - 2 = r$, which is the same as before. So, that's already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to another root, but not covered in Subcase 2c? For example, suppose that one root is 2 (which maps to itself) and the other root is something else. Wait, but if one root is 2, then the other root must satisfy that $c^2 - 2$ is a root. Since 2 is already a root, then the other root must be either 2 or something else. If the other root is 2, then it's a repeated root. If the other root is something else, say $s$, then $s^2 - 2$ must be a root. Since the roots are 2 and $s$, then $s^2 - 2$ must be either 2 or $s$. If $s^2 - 2 = 2$, then $s^2 = 4$, so $s = \pm 2$. If $s = 2$, then repeated root. If $s = -2$, then we have roots 2 and -2, which is the quadratic $x^2 - 4$, which is already considered. If $s^2 - 2 = s$, then $s^2 - s - 2 = 0$, which gives $s = 2$ or $s = -1$. If $s = 2$, repeated root. If $s = -1$, then roots are 2 and -1, which is quadratic $x^2 - x - 2$, already considered. Therefore, no new quadratics here.

Similarly, if one root is -1, then the other root must satisfy $s^2 - 2$ is a root. If the other root is -1, then repeated root. If the other root is something else, say $s$, then $s^2 - 2$ must be either -1 or $s$. If $s^2 - 2 = -1$, then $s^2 = 1$, so $s = \pm 1$. If $s = 1$, then roots are -1 and 1, quadratic $x^2 - 1$, already considered. If $s = -1$, repeated root. If $s^2 - 2 = s$, then $s^2 - s - 2 = 0$, which gives $s = 2$ or $s = -1$. If $s = 2$, then roots are -1 and 2, quadratic $x^2 - x - 2$, already considered. If $s = -1$, repeated root. Therefore, again, no new quadratics.

Therefore, all possibilities seem to be covered. Therefore, total of 6 quadratics. Wait, but let me check again:

Case 1: 2 quadratics.

Case 2a: 1 quadratic.

Case 2b: 1 quadratic.

Case 2c: 2 quadratics.

Total: 2 + 1 + 1 + 2 = 6.

But let me check if there are any other quadratics. For example, suppose that we have a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? Wait, that's exactly Subcase 2b, which gives us the quadratic $x^2 + x - 1$. So that's already considered.

Alternatively, could there be quadratics where both roots map to each other in some other way? For example, if $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots? But that's the same as Subcase 2b. So, no.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That's Case 2a, which gives roots 2 and -1. So that's already considered.

Alternatively, could there be quadratics where one root maps to itself and the other maps to another root, but not covered in Case 2c? For example, if one root is 2 (maps to itself) and the other root is something else. But we already considered that in Case 2c. Similarly for -1.

Therefore, I think we have all the quadratics. Therefore, total of 6 quadratics. However, let me check each quadratic to ensure they satisfy the condition.

First, Case 1:

1. $x^2 - 4x + 4$: roots 2, 2. For root 2: 2^2 - 2 = 2, which is a root. Good.

2. $x^2 + 2x + 1$: roots -1, -1. For root -1: (-1)^2 - 2 = -1, which is a root. Good.

Case 2a:

3. $x^2 - x - 2$: roots 2, -1. For root 2: 2^2 - 2 = 2, which is a root. For root -1: (-1)^2 - 2 = -1, which is a root. Good.

Case 2b:

4. $x^2 + x - 1$: roots $[-1 \pm \sqrt{5}]/2$. Let me denote them as $r_1$ and $r_2$. For $r_1$, $r_1^2 - 2 = r_2$, which is a root. Similarly for $r_2$, $r_2^2 - 2 = r_1$. Good.

Case 2c:

5. $x^2 - 4$: roots 2, -2. For root 2: 2^2 - 2 = 2, which is a root. For root -2: (-2)^2 - 2 = 2, which is a root. Good.

6. $x^2 - 1$: roots 1, -1. For root 1: 1^2 - 2 = -1, which is a root. For root -1: (-1)^2 - 2 = -1, which is a root. Good.

Therefore, all six quadratics satisfy the condition. Therefore, the total number is 6. However, let me check if there are any other quadratics that I might have missed.

Wait, let me think again. For example, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots. Wait, that's exactly the quadratic $x^2 + x - 1$, which we already considered. So, no new quadratics there.

Alternatively, could there be quadratics where one root maps to another root, and that root maps to another root, but not forming a cycle? But since there are only two roots, if $r^2 - 2 = s$ and $s^2 - 2 = r$, then applying the transformation twice gives $r^4 - 4r^2 + 2 = r$, which is the quartic equation we solved earlier. Therefore, the roots of that quartic are already considered. Therefore, no new quadratics.

Alternatively, could there be quadratics where one root maps to itself and the other maps to itself? That's Case 2a, which gives roots 2 and -1. So that's already considered.

Therefore, I think we have all the quadratics. Therefore, the answer is 6. However, let me check once more for possible duplicates or errors.

Wait, but let me check if the quadratic $x^2 + x - 1$ is distinct from the others. Yes, because its roots are different from the others. Similarly, $x^2 - 4$ and $x^2 - 1$ are distinct. The quadratics in Case 1 are distinct from the others. The quadratic $x^2 - x - 2$ is distinct. Therefore, total of 6 quadratics.

But wait, hold on. Let me check if there are quadratics with different coefficients but same roots. For example, if two different quadratics have the same roots, but since we are considering monic quadratics (since the problem says "quadratic equations of the form $x^2 + ax + b = 0$"), then each set of roots corresponds to exactly one quadratic. Therefore, each distinct set of roots gives a unique quadratic. Therefore, since we have 6 distinct quadratics, the answer is 6.

But let me check once more if there are other quadratics. For example, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots. Wait, but that's exactly the quadratic $x^2 + x - 1$, which we already considered. Therefore, no.

Alternatively, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with $r$ and $s$ being different from the ones in $x^2 + x - 1$. But solving $r^2 - 2 = s$ and $s^2 - 2 = r$ leads to the quartic equation which factors into (r + 1)(r - 2)(r^2 + r - 1) = 0. Therefore, the roots are -1, 2, and the roots of $r^2 + r - 1 = 0$. Therefore, the only possible pairs of roots are:

- (-1, -1) [repeated root]

- (2, 2) [repeated root]

- (-1, 2) [distinct roots]

- (roots of $r^2 + r - 1 = 0$) [distinct roots]

- (-1, -2) [from Subcase 2c]

- (1, -1) [from Subcase 2c]

Wait, but how does (-1, -2) come into play? Let me think. If we have roots -1 and -2, then for root -1: (-1)^2 - 2 = -1, which is a root. For root -2: (-2)^2 - 2 = 2, which is not a root. Wait, hold on! Wait, earlier I thought that quadratic $x^2 - 4$ has roots 2 and -2. Let me check again. If the quadratic is $x^2 - 4$, then the roots are 2 and -2. For root 2: 2^2 - 2 = 2, which is a root. For root -2: (-2)^2 - 2 = 4 - 2 = 2, which is a root. Therefore, yes, that works. But if we have roots -1 and -2, then for root -2, $c^2 - 2 = 2$, which is not a root unless 2 is a root. But if the quadratic is $x^2 - 4$, then 2 is a root. Therefore, the roots are 2 and -2. Therefore, the quadratic $x^2 - 4$ is valid. However, if we have roots -1 and -2, then the quadratic would be $x^2 - (-1 + (-2))x + (-1)(-2) = x^2 + 3x + 2$. Let me check if this quadratic satisfies the condition. For root -1: (-1)^2 - 2 = -1, which is a root. For root -2: (-2)^2 - 2 = 2, which is not a root of $x^2 + 3x + 2$. Therefore, this quadratic does not satisfy the condition. Therefore, my earlier reasoning was wrong. Wait, how did I get the quadratic $x^2 - 4$?

Earlier, in Subcase 2c, when I considered $s = 2$ and $r^2 - 2 = 2$, leading to $r = \pm 2$. If $r = -2$, then the roots are -2 and 2, which is the quadratic $x^2 - 4$. Therefore, that quadratic is valid. However, if I have roots -1 and -2, that quadratic is not valid. Therefore, my mistake earlier was thinking that if $s = -1$, then $r^2 - 2 = -1$ gives $r = \pm 1$, leading to roots 1 and -1. Therefore, quadratic $x^2 - 1$. Therefore, the quadratic $x^2 - 1$ has roots 1 and -1. For root 1: 1^2 - 2 = -1, which is a root. For root -1: (-1)^2 - 2 = -1, which is a root. Therefore, that quadratic is valid. However, if I have roots -1 and -2, that quadratic is invalid. Therefore, my earlier reasoning was correct in that only quadratics with roots 2 and -2, and 1 and -1 are valid. Therefore, the quadratics $x^2 - 4$ and $x^2 - 1$ are valid. Therefore, my previous conclusion of 6 quadratics is correct.

Therefore, the total number of such quadratics is 6. However, let me check once more if there are any other quadratics.

Wait, let me check if the quadratic $x^2 + x - 1$ is the only one from Subcase 2b. Yes, because the roots are from the equation $x^2 + x - 1 = 0$, which is unique. Therefore, that's one quadratic.

Therefore, total of 6 quadratics. Therefore, the answer is 6. However, let me check once more for possible mistakes.

Wait, but let me check if the quadratic $x^2 + x - 1$ is the only one from Subcase 2b. Yes, because the quartic equation had roots -1, 2, and the roots of $x^2 + x - 1 = 0$. Therefore, the only pair of roots that are not repeated and not involving -1 or 2 is the roots of $x^2 + x - 1 = 0$. Therefore, that's one quadratic.

Therefore, total of 6 quadratics. Therefore, the answer is 6. However, let me check if the problem says "quadratic equations of the form $x^2 + ax + b = 0$", so they are monic quadratics. Therefore, each distinct pair of roots gives a unique quadratic. Therefore, 6 quadratics. Therefore, the answer is 6.

But wait, let me check once more for possible duplicates. For example, is $x^2 - x - 2$ the same as any other quadratic? No. Similarly, $x^2 + x - 1$ is unique. The quadratics in Case 1 are unique. The quadratics in Case 2c are unique. Therefore, total of 6.

But let me check if there are other quadratics where, for example, both roots map to each other in some other way. For example, suppose that $r^2 - 2 = s$ and $s^2 - 2 = r$, but with different roots. Wait, but that's exactly the quadratic $x^2 + x - 1$, which we already considered. Therefore, no.

Alternatively, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with $r$ and $s$ being different from the ones in $x^2 + x - 1$. But solving $r^2 - 2 = s$ and $s^2 - 2 = r$ leads to the quartic equation, which only has those roots. Therefore, no.

Therefore, I think the answer is 6. However, I need to check if the problem counts quadratics with the same roots but different coefficients. But since the problem says "quadratic equations of the form $x^2 + ax + b = 0$", each distinct pair of roots gives a unique quadratic. Therefore, 6 quadratics.

But wait, let me check once more. Let me list all six quadratics:

1. $x^2 - 4x + 4$ (a = -4, b = 4)

2. $x^2 + 2x + 1$ (a = 2, b = 1)

3. $x^2 - x - 2$ (a = -1, b = -2)

4. $x^2 + x - 1$ (a = 1, b = -1)

5. $x^2 - 4$ (a = 0, b = -4)

6. $x^2 - 1$ (a = 0, b = -1)

Yes, all these have different coefficients (a, b). Therefore, they are distinct equations. Therefore, the total number is 6.

But wait, hold on. The quadratic $x^2 - 4$ has a = 0, b = -4. The quadratic $x^2 - 1$ has a = 0, b = -1. So they are different. Similarly, the others have different a and b. Therefore, 6 distinct quadratics.

Therefore, the answer is 6. However, I need to check if there are any other quadratics that I might have missed. For example, suppose that there is a quadratic with roots $r$ and $s$ where $r^2 - 2 = s$ and $s^2 - 2 = r$, but with $r$ and $s$ being complex numbers? Wait, but the problem doesn't specify that the roots have to be real. However, the problem says "quadratic equations of the form $x^2 + ax + b = 0$", and doesn't specify that the coefficients a and b are real numbers. Wait, but usually, in such problems, unless specified otherwise, we assume that a and b are real numbers. However, even if they are complex, the problem might still have solutions. However, the condition is that whenever c is a root, c^2 - 2 is also a root. If the roots are complex, then c^2 - 2 might not be a root unless specific conditions are met. However, since the problem doesn't specify, but given that the answer is likely to be an integer, and given that the previous cases have real roots, maybe the problem assumes real coefficients and real roots? Or maybe complex roots are allowed?

Wait, but if complex roots are allowed, then there might be more quadratics. However, since the problem is likely from a competition or textbook problem, it's more probable that they are considering real coefficients and real roots. However, let me check.

Suppose that we allow complex roots. Then, for example, the quadratic $x^2 + x - 1 = 0$ has complex roots? No, wait, discriminant is $1 + 4 = 5$, which is positive, so real roots. The quadratic $x^2 + x - 1$ has real roots. The quadratic $x^2 - x - 2$ has real roots. The quadratics in Case 1 have real roots. The quadratics in Case 2c have real roots. Therefore, all quadratics considered so far have real roots. Therefore, even if complex roots are allowed, there are no additional quadratics because the other roots would be complex and would not satisfy the condition unless they are closed under the transformation $c \mapsto c^2 - 2$. However, since the problem didn't specify, but given that all the quadratics we found have real roots, and the answer is 6, I think that is the intended answer.

Therefore, I think the number of such quadratic equations is 6. Therefore, the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To solve the problem, we are asked to find the number of **monic quadratic equations** of the form:

$$
x^2 + ax + b = 0
$$

such that **whenever $ c $ is a root**, then $ c^2 - 2 $ is also a root.

---

### Step 1: Understanding the Condition

Let the roots of the quadratic be $ r_1 $ and $ r_2 $. The condition implies that for each root $ c $, the value $ c^2 - 2 $ must also be a root. Since the quadratic has only two roots, this condition must be satisfied for both roots.

This leads to the following possibilities:

- If $ r_1 = r_2 $, then the quadratic has a **repeated root**. In this case, the condition implies that $ r^2 - 2 = r $, which gives a quadratic equation in $ r $:
  $$
  r^2 - r - 2 = 0 \Rightarrow (r - 2)(r + 1) = 0
  $$
  So the repeated roots are $ r = 2 $ or $ r = -1 $, leading to two quadratics:
  - $ (x - 2)^2 = x^2 - 4x + 4 $
  - $ (x + 1)^2 = x^2 + 2x + 1 $

- If $ r_1 \ne r_2 $, then the roots must satisfy the condition that $ r_1^2 - 2 = r_2 $ and $ r_2^2 - 2 = r_1 $. This leads to the system:
  $$
  r_1^2 - 2 = r_2 \quad \text{and} \quad r_2^2 - 2 = r_1
  $$
  Substituting $ r_2 = r_1^2 - 2 $ into the second equation gives:
  $$
  (r_1^2 - 2)^2 - 2 = r_1 \Rightarrow r_1^4 - 4r_1^2 - r_1 + 2 = 0
  $$
  Factoring this quartic yields:
  $$
  (r + 1)(r - 2)(r^2 + r - 1) = 0
  $$
  The roots are:
  - $ r = -1 $, $ r = 2 $, and the roots of $ r^2 + r - 1 = 0 $, which are:
    $$
    r = \frac{-1 \pm \sqrt{5}}{2}
    $$

  From this, we can form the following distinct quadratics:

  - Roots $ r = 2 $ and $ r = -1 $: $ x^2 - x - 2 $
  - Roots $ r = \frac{-1 + \sqrt{5}}{2} $ and $ r = \frac{-1 - \sqrt{5}}{2} $: $ x^2 + x - 1 $

  Additionally, we consider the case where one root maps to itself and the other maps to the first root. For example:

  - If $ r = 2 $, then $ r^2 - 2 = 2 $, so the other root must be $ -2 $, leading to the quadratic $ x^2 - 4 $
  - If $ r = -1 $, then $ r^2 - 2 = -1 $, so the other root must be $ 1 $, leading to the quadratic $ x^2 - 1 $

---

### Step 2: List All Valid Quadratics

From the above, we have the following **six distinct monic quadratic equations** that satisfy the condition:

1. $ x^2 - 4x + 4 $ (repeated root 2)
2. $ x^2 + 2x + 1 $ (repeated root -1)
3. $ x^2 - x - 2 $ (roots 2 and -1)
4. $ x^2 + x - 1 $ (roots $ \frac{-1 \pm \sqrt{5}}{2} $)
5. $ x^2 - 4 $ (roots 2 and -2)
6. $ x^2 - 1 $ (roots 1 and -1)

Each of these quadratics has been verified to satisfy the condition that if $ c $ is a root, then $ c^2 - 2 $ is also a root.

---

### Final Answer

$$
\boxed{6}
$$

Solution:
Let the roots be $r$ and $s$ (not necessarily real).  We take the cases where $r = s$ and $r \neq s.$

Case 1: $r = s.$

Since $r$ is the only root, we must have $r^2 - 2 = r.$  Then $r^2 - r - 2 = 0,$ which factors as $(r - 2)(r + 1) = 0,$ so $r = 2$ or $r = -1.$  This leads to the quadratics $x^2 - 4x + 4$ and $x^2 + 2x + 1.$

Case 2: $r \neq s.$

Each of $r^2 - 2$ and $s^2 - 2$ must be equal to $r$ or $s.$  We have three cases:

(i) $r^2 - 2 = r$ and $s^2 - 2 = s.$

(ii) $r^2 - 2 = s$ and $s^2 - 2 = r.$

(iii) $r^2 - 2 = s^2 - 2 = r$.

In case (i), as seen from Case $r,$ $s \in \{2,-1\}.$  This leads to the quadratic $(x - 2)(x + 1) = x^2 - x - 2.$

In case (ii), $r^2 - 2 = s$ and $s^2 - 2 = r.$  Subtracting these equations, we get
\[r^2 - s^2 = s - r.\]Then $(r - s)(r + s) = s - r.$  Since $r - s \neq 0,$ we can divide both sides by $r - s,$ to get $r + s = -1.$  Adding the equations $r^2 - 2 = s$ and $s^2 - 2 = r,$ we get
\[r^2 + s^2 - 4 = r + s = -1,\]so $r^2 + s^2 = 3.$  Squaring the equation $r + s = -1,$ we get $r^2 + 2rs + s^2 = 1,$ so $2rs = -2,$ or $rs = -1.$  Thus, $r$ and $s$ are the roots of $x^2 + x - 1.$

In case (iii), $r^2 - 2 = s^2 - 2 = r.$  Then $r^2 - r - 2 = 0,$ so $r = 2$ or $r = -1.$

If $r = 2,$ then $s^2 = 4,$ so $s = -2.$  (We are assuming that $r \neq s.$)  This leads to the quadratic $(x - 2)(x + 2) = x^2 - 4.$

If $r = -1$, then $s^2 = 1,$ so $s = 1.$  This leads to the quadratic $(x + 1)(x - 1) = x^2 - 1.$

Thus, there are $\boxed{6}$ quadratic equations that work, namely $x^2 - 4x + 4,$ $x^2 + 2x + 1,$ $x^2 - x - 2,$ $x^2 + x - 1,$ $x^2 - 4,$ and $x^2 - 1.$