<s>[INST] 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
 [/INST] Hint
You have missed $$(-1)^{\text{even number}}=1$$
Now if $x^2-5x+5=-1,x=?$
Which values of $x$ make the exponent even?
</s>