Number of Integers Satisfying Exponential Equality

An equation of the form $A^B = 1$ holds in three cases:

1. $B = 0$ and $A \ne 0$: $x + 1 = 0 \implies x = -1$ Check $A$: $(-1)^2 - 5(-1) + 7 = 1 + 5 + 7 = 13 \ne 0$. So $x = -1$ is a solution.

2. $A = 1$: $x^2 - 5x + 7 = 1 \implies x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0 \implies x = 2, 3$ Both are integers.

3. $A = -1$ and $B$ is an even integer: $x^2 - 5x + 7 = -1 \implies x^2 - 5x + 8 = 0$ Discriminant = $25 - 32 < 0$, no real solutions.

Thus, the solutions are $x = -1, 2, 3$. The total number of integers is 3.