<s>[INST] System of Equations involving exponents I would like to solve the system:

$$2^{x+y^2}+2^{x^2+y}=8 $$
  $$\sqrt x+ \sqrt y =2 $$

I have come up with a solution that is not rigorous, but somehow it gives the right solution:
Expanding the second equation:
$$x+y +2\sqrt {xy}=4$$
Squaring again
$$(x+y-4)^2 =4xy$$
$$x^2+y^2-2xy-8x-8y+16=0$$
$$2^{ x^2+y^2-2xy-8x-8y+16}=1$$
Rewriting the expression in the form:
$$2^{x^2+y^2+x+y} =2^{9x+9y+2xy-16}$$
Now I let $a(x,y)=2^{x^2+y}$ and $b(x,y)=2^{y^2 +x}$, and let $f(x,y)$ be the right hand side of the above equation, then we have:
$$a+b=8$$
$$ab=f(x,y)$$
Without justification, I use quadratic formula by treating $f(x,y)$ like a coefficient, I get:
$$a(x,y)=4 \pm \sqrt{16-f(x,y)}$$
$$b(x,y)=4 \mp \sqrt{16-f(x,y)}$$
Since $a(x,y)=a(y,x)$ based on the expression given from quadratic formula, 
$$2^{x+y^2}=2^{x^2+y}$$
Factorize:
$$(x-y)(x+y-1)=0$$
$x=y$ or $x+y=1$
At this point it is easy to verify that $x=y=1$ is the only solution.

Question: Is my method of applying quadratic formula to a function valid? What are the only ways to solve this question?

 [/INST] By quadratic-arithmetic inequality, we have $$(x+y)\geq \frac{(\sqrt{x}+\sqrt{y})^2}{2}=2.$$ 
Similarly, $$(x^2+y^2)\geq\frac{(x+y)^2}{2}\geq 2.$$
Also, by arithmetic-geometric inequality, we have  $$2^{x+y^2}+2^{x^2+y}\geq 2\sqrt{2^{x+y^2}2^{x^2+y^2}}\geq 2\sqrt{2^{4}}=8.$$ Here equality holds, which is possible only if $\sqrt{x}=\sqrt{y}$. 
Therefore, $x=y=1$ is the only solution.
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