The number of integers that satisfy the equality $(x^2 - 5x + 7)^{x + 1} = 1$ is

To solve $(x^2 - 5x + 7)^{x+1} = 1$, we need to understand when an expression raised to a power equals 1.

There are three main cases where $a^b = 1$:

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When the base equals 1, any exponent will give us 1.

$x^2 - 5x + 7 = 1$

$x^2 - 5x + 6 = 0$

We need two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3.

$(x - 2)(x - 3) = 0$

Therefore: $x = 2$ or $x = 3$

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When the exponent is 0, we get $a^0 = 1$ (provided $a

eq 0$).

$x + 1 = 0$

$x = -1$

Check that base ≠ 0:

When $x = -1$: $(-1)^2 - 5(-1) + 7 = 1 + 5 + 7 = 13

eq 0$

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When base = -1 and exponent is even, we get $(-1)^{\text{even}} = 1$.

$x^2 - 5x + 7 = -1$

$x^2 - 5x + 8 = 0$

Using the discriminant: $\Delta = (-5)^2 - 4(1)(8) = 25 - 32 = -7 < 0$

Since the discriminant is negative, there are no real solutions for this case.

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The integer solutions are: $x = -1, 2, 3$

Therefore, the number of integers that satisfy the equation is 3.

Key Learning: Remember the three cases where $a^b = 1$:

1. $a = 1$ (any exponent)
2. $b = 0$ (base ≠ 0)
3. $a = -1$ and $b$ is even

This systematic approach ensures we don't miss any solutions!