Polynomial Exponent

There are three cases to check when solving this problem: when the base is equal to $1,$ when the exponent is equal to $0$ and the base is not equal to $0,$ and when the base is equal to $-1$ and the exponent is divisible by $2.$

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$\textbf{Case 1:}$ $x^2-7x+11=1$

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

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

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

This shows that $x=2$ and $x=5$ are solutions.

$\textbf{Case 2:}$ $x^2-6x-7=0$

$x^2-6x-7=0$

$\Rightarrow(x+1)(x-7)=0$

Checking $x=-1$ and $x=7$ in the base, we find that neither of those cause the base to be equal to $0,$ so $x=-1$ and $x=7$ can be accepted as solutions.

$\textbf{Case 3:}$ $x^2-7x+11=-1, x\equiv1\text{ mod }2$

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

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

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

Checking $x=3$ and $x=4$ in the exponent, we find that $x=3$ causes the exponent to be divisible by $2$ but $x=4$ does not. Thus, we can only accept $x=3$ as a solution.

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Our solutions are $x\in\{-1,2,3,5,7\}.$ The sum of the solutions is $\boxed{16}$

The condition of the problem is satisfied if one of the following cases hold true: $$ $\text{Case 1}$ : $x^2 -7x+11 = 1$ and the exponent can be any number. This yields $x=5,2$ $$ $\text{Case 2}$ : $x^2-7x+11 = -1$ and the exponent must be even. This yields $x= 3$ $$

$\text{Case 3}$ : $x^2-6x-7=0$ This yields $x= 7, -1$ $$

So the sum of all values of $x$ is $5+2+3+7-1 = \boxed{16}$

Case 1: $x^2-6x-7=0$

$x^2-6x-7=0$

$(x-7)(x+1)=0$

$x=7,-1$

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

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

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

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

$x=5,2$

Case 3: $x^2-7x+11=1$ and $x^2-6x-7=2k$

As $x^2-6x-7=(x-7)(x+1), (x-7)$ and $(x+1)$ have the same parity $\therefore$ $x^2-6x-7=2k$ only for odd $x$

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

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

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

Then odd $x$ and solution of $x^2-7x+11=-1$ is $x=3$

The sum of all values for $x$ is $7-1+5+2+3=\boxed{16}$