The number of integer solutions of the equation $\left(x^{2}-10\right)^{\left(x^{2}-3 x-10\right)}=1$ is

When we have an equation of the form $a^b = 1$, there are exactly three scenarios where this can happen:

Case 1: Base equals 1 (any exponent works)

Case 2: Base equals -1 and exponent is even

Case 3: Base is non-zero and exponent equals 0

Let us identify our components:

Base: $a = x^2 - 10$

Exponent: $b = x^2 - 3x - 10$

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Case 1: Base = 1

$x^2 - 10 = 1$

$x^2 = 11$

$x = \pm\sqrt{11}$

Since $\sqrt{11}$ is not an integer, Case 1 gives no integer solutions.

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Case 2: Base = -1 and Exponent is Even

$x^2 - 10 = -1$

$x^2 = 9$

$x = \pm 3$

Now we need to check if the exponent is even for both values:

For $x = 3$:

$b = 3^2 - 3(3) - 10 = 9 - 9 - 10 = -10$

Since $-10$ is even, $(-1)^{-10} = 1$

For $x = -3$:

$b = (-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8$

Since $8$ is even, $(-1)^8 = 1$

Case 2 gives us solutions: $x = 3$ and $x = -3$

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Case 3: Non-zero Base and Exponent = 0

$x^2 - 3x - 10 = 0$

We'll factor this quadratic:

$x^2 - 3x - 10 = (x - 5)(x + 2) = 0$

So $x = 5$ or $x = -2$

Let us verify the base is non-zero for both:

For $x = 5$:

$a = 5^2 - 10 = 25 - 10 = 15 \neq 0$

For $x = -2$:

$a = (-2)^2 - 10 = 4 - 10 = -6 \neq 0$

Case 3 gives us solutions: $x = 5$ and $x = -2$

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Combining all cases, our integer solutions are:

$x = 3, -3, 5, -2$

Total number of integer solutions = 4