Sum of all solutions of $(x^2+5x+5)^{(x-3)(x-7)}=1$ 
Find the sum of all solutions of $(x^2+5x+5)^{(x-3)(x-7)}=1$

An obvious approach is to consider the case where the exponent is $0$ which yields the solutions $3,7$ and then the case where the base is $1$ which yields the solutions $-1,-4$ and then finally to consider the case where base is $-1$ and the exponent is even,which yields the solution $-3$. But the thing is,how do we know these are the only solutions.I don't see any reason these should be the only solutions.Why can't there be complex solutions,or perhaps even more real solutions.Graphing might give us insight,but I don't have any good graphing software at the moment.
I have also thought about using Vieta's formulas,but that seems to complicate things.
Any help,hint or prod in the right direction will be appreciated.Also,is there any way we can know about how the graph of the function will behave without actually graphing the function?
 A: I will explain how to find the real solutions. You already found all of them. I don't think there are complex solutions, but I had not managed to prove it.
Let $f(x) = (x^2 + 5x + 5)^{(x - 3)(x - 7)}$, and $\alpha = -\frac{5 + \sqrt 5}2,~\beta = \frac{-5 + \sqrt 5}2$, that is, the solutions to $x^2 + 5x + 5$. The domain of the function $f$ is
$$\operatorname{dom} f = (-\infty,~\alpha)\cup\{x \in (\alpha,~\beta) : (x-3)(x-7) \in \mathbb Z\}\cup(\beta,~+\infty).$$
Use the fact that
$$x = \begin{cases}
\quad\exp(\ln(x)) & x > 0\\
-\exp(\ln(-x)) & x < 0
\end{cases}$$
In your case, we have:
$$f(x) = \begin{cases}
\quad\exp((x - 3)(x - 7)\ln(x^2 + 5x + 5)) = \kappa(x) & x \in (-\infty,~\alpha) \cup (\beta,~+\infty)\\
-\exp((x - 3)(x - 7)\ln(-x^2 - 5x - 5)) = \chi(x) & x \in (\alpha,~\beta)
\end{cases}.$$
You can verify by substitution that $\alpha$ and $\beta$ are not solutions to the original equation.
Therefore we have to solve
$$\begin{cases}
x \in (-\infty,~\alpha) \cup (\beta,~+\infty)\\
\kappa(x) = 1
\end{cases} \quad\lor\quad \begin{cases}x \in (\alpha,~\beta)\\
\chi(x) = 1
\end{cases}$$
The second system has no solution in the real numbers. As for the first one,
$$\begin{cases}
x \in (-\infty,~\alpha) \cup (\beta,~+\infty)\\
x = 3 \lor x = 7 \lor x \in \{-4, -3, -2, -1\}
\end{cases}$$
Which gives
$$x \in \{-4, -1, 3, 7\}.$$
We didn't however consider the case in which the base is negative and the exponent is a number. We are interested in the case in which the base is equal to $-1$ and the exponent is even. This gives, as you mentioned in the question, $x = -3$.
Therefore all the real solutions are
$$x \in \{-4, -3, -1, 3, 7\}$$
and their sum is $2$.
A: If $a^x=1$ where $a$  is real
either $x=0$
or $a=1$
or $a=-1$ and $x$ is even
A: Hint:
Use logarithm
$(x-3)(x-7)\log{(x^2+5x+5)}=0$
Logarithm definition can extend to negative numbers too.
A: Taking "$\ln$" on both sides we get,
$(x-3)(x-7)\ln(x^2+5x+5)=\ln 1$  [Assuming the term inside $\ln$ is positive]
$\implies (x-3)(x-7)\ln(x^2+5x+5)= 0$ 
Now there may arise 3 cases....
The two are easy to find... $x=3,7$
Now what can we do with the complex one,
$\ln(x^2+5x+5)=0$
$\implies x^2+5x +5 =1$
$\implies x^2+5x +4 =0 $
Oops!!!! We get the same result....
