Sum of all real numbers $x$ such that $(\text{A quadratic})^\text{Another quadratic}=1$. 
What is the sum of all real numbers $x$ such that
  $(x^2-5x+5)^{(x^2-7x+12)}=1$?

So I know that $x^0=1$ and $1^x=1$. So, I can solve for them and find $x$, and add them up. 
Solving $x^2-7x+12=0$ for $x^0=1$ gives $x=3, 4$.
Solving $x^2-5x+5=1$ for $1^x=1$ gives $x=1, 4$.
Adding them up gives $1+3+4=8$.
This is wrong. What did I do wrong? Did I miss a case? If so, what case have I missed?
Thanks!
Max0815

Three cases:
$$x^0=1$$
$$1^x=1$$
$$(-1)^{\text{even #}}=1$$
Substituting into the original equation gives $$x=3, 4$$ $$x=1, 4$$ $$x=2, 3$$
We don't add in the solutions already accounted for, namely, 3 and 4.
$\sum=3+4+1+2=10$.
