Is it possible that $2^{2A}+2^{2B}$ is a square number? 
Let A and B be two positive integers greater than $0$. Is it possible that $2^{2A}+2^{2B}$ is a square number?

I am having trouble with this exercise because I get the feeling the answer is no, but I cannot elaborate on the proof. So far what I thought was to assume that there is some integer $C>0$ such that  $2^{2A}+2^{2B}=C^2$. Then
$$(2^A+2^B)^2=C^2+2^{A+B+1}$$
I was trying to see if the previous expression could hold a contradiction but I got stuck. All I could find is that $C$ needs to be an even number but that doesn't seem to get me anywhere. I'd appreciate any help.
Thanks in advance!

Just do it.
Assume without loss of generality that $A \le B$ so
$2^{2A} + 2^{2B}=$
$2^{2A} (1 + 2^{2B-2A})=$
$(2^A)^2 [1 + 2^{2B-2A}]=$
$(2^A)^2 [(2^{B-A})^2 + 1]$.
So if that is a perfect square then we must have $(2^{B-A})^2 + 1$ being a perfect square.
But $(2^{B-A})^2$ is a perfect square so we have two consecutive perfect squares.  It should be easy to convince yourself that the only time that ever occurs is $0^2$ and $1^2$.  (Proof as addendum).
So the only way this can happen is if $(2^{B-A})^2 = 0$ and $(2^{B-A})^2 + 1=1$.
But $2^{B-A} = 0$ is not possible.
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Addendume:  Then only two consecutive squares are $0$ and $1$.
Proof:  Suppose $m^2 = n^2 + 1$.  where $m,n$ are non-negative integers. $n^2 < m^2 = n^2 + 1 \le n^2 + 2n + 1= (n+1)^2$ so $n < m \le m+1$.  But the only integers between $n$ (exclusive) and $n+1$ (inclusive) is $n+1$ so $m = n+1$.  And so $n^2 + 1 = m^2 = (n+1) = n^2 + 2n + 1$ so $2n = 0$ and $n = 0$ and $m =1$.
