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

Question

Accepted Answer

To solve the equation $$(x^2 - 10)^{(x^2 - 3x - 10)} = 1$$, we analyze the cases where the exponential equation equals $$1$$:

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### Case Analysis:

#### Case 1: $$b = x^2 - 3x - 10 = 0$$  
If $$b = 0$$, then $$a^0 = 1$$ for any $$a 
eq 0$$.  
Solve $$x^2 - 3x - 10 = 0$$:  
$$
(x - 5)(x + 2) = 0
$$  
Thus, $$x = 5$$ or $$x = -2$$.

#### Case 2: $$a = x^2 - 10 = 1$$  
If $$a = 1$$, then $$1^b = 1$$ for any $$b$$.  
Solve $$x^2 - 10 = 1$$:  
$$
x^2 = 11
$$  
Since $$\sqrt{11}$$ is not an integer, there are no solutions in this case.

#### Case 3: $$a = x^2 - 10 = -1 	ext{ and } b 	ext{ is even}$$  
If $$a = -1$$, then $$(-1)^b = 1$$ only if $$b 	ext{ is even}$$.  
Solve $$x^2 - 10 = -1$$:  
$$
x^2 = 9
$$  
Thus, $$x = \pm3$$.

For both values of $$x = \pm3$$:  
- Calculate $$b = x^2 - 3x - 10$$:  
    - For $$x = +3$$: $$b = (3)^2 - (3)(3) - 10 = -10 \, (	ext{even})$$.  
    - For $$x = -3$$: $$b = (-3)^2 - (-3)(3) - 10 = +8 \, (	ext{even})$$.

Thus, both values satisfy the condition.

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### Total Solutions:  
From the above cases:  
- Case 1 gives two solutions: $$x = 5, x = -2$$.  
- Case 3 gives two solutions: $$x = +3, x = -3$$.

In total, there are **4 integer solutions**: $$x = \{-3, +3, -2, +5\}$$.

Thus, the correct answer is **4**.