How many distinct positive integer-valued solutions exist to the equation $\left(x^{2}-7 x+11\right)^{\left(x^{2}-13 x+42\right)}=1 ?$

When does $a^b = 1$? There are exactly three situations:

The base equals 1: $a = 1$ (exponent can be anything)

The base equals -1 and the exponent is even: $a = -1$ and $b$ is even

The exponent equals 0: $b = 0$ (base cannot be 0)

Let's apply this to our equation: $(x^2 - 7x + 11)^{(x^2 - 13x + 42)} = 1$

---

We set the base equal to 1:

$x^2 - 7x + 11 = 1$

$x^2 - 7x + 10 = 0$

We need two numbers that multiply to 10 and add to -7.

Those numbers are -2 and -5.

$(x - 2)(x - 5) = 0$

Solutions: $x = 2$ and $x = 5$

---

We set the base equal to -1:

$x^2 - 7x + 11 = -1$

$x^2 - 7x + 12 = 0$

We need two numbers that multiply to 12 and add to -7.

Those numbers are -3 and -4.

$(x - 3)(x - 4) = 0$

Potential solutions: $x = 3$ and $x = 4$

For $(-1)^{\text{exponent}} = 1$, the exponent must be even.

When $x = 3$: Exponent = $3^2 - 13(3) + 42 = 9 - 39 + 42 = 12$ (even)

When $x = 4$: Exponent = $4^2 - 13(4) + 42 = 16 - 52 + 42 = 6$ (even)

Valid solutions: $x = 3$ and $x = 4$

---

We set the exponent equal to 0:

$x^2 - 13x + 42 = 0$

We need two numbers that multiply to 42 and add to -13.

Those numbers are -6 and -7.

$(x - 6)(x - 7) = 0$

Potential solutions: $x = 6$ and $x = 7$

We need to ensure the base isn't 0 (since $0^0$ is undefined).

When $x = 6$: Base = $6^2 - 7(6) + 11 = 36 - 42 + 11 = 5

eq 0$

When $x = 7$: Base = $7^2 - 7(7) + 11 = 49 - 49 + 11 = 11

eq 0$

Valid solutions: $x = 6$ and $x = 7$

---

All distinct positive integer solutions:

From Case 1: $x = 2, 5$

From Case 2: $x = 3, 4$

From Case 3: $x = 6, 7$

Total count: 6 distinct solutions

When solving exponential equations equal to 1, we always check all three cases systematically. This approach ensures we don't miss any solutions.