<s>[INST] Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$

I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question... 
 [/INST] Set $a=2^x$, $b=2^y$, then the problem is equivalent to finding the range of $a+b$ where $a,b>0$ and
$$a^2+b^2=2a+2b$$
Without loss of generality $a\ge b$ and we can make the substitution $c=a+b$, $d=a-b$, so now $c>d\ge0$, and we require
\begin{align*}\left(\frac{c+d}2\right)^2+\left(\frac{c-d}2\right)^2&=(c+d)+(c-d)\\
c^2+2cd+d^2+c^2-2cd+d^2&=8c\\
c^2+d^2&=4c\end{align*}
Observe that
$$c^2\le c^2+d^2=4c\Longleftrightarrow c(c-4)\le0\Longleftrightarrow c\le4$$
and
$$2c^2>c^2+d^2=4c\Longleftrightarrow c(c-2)>0\Longleftrightarrow c>2$$
It's easy to see all such values can be attained and the answer is the interval $(2,4]$.
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