An evenly even number is a number in the form of 2^m , where m is a positive integer.
Prove that it is impossible the sum of two evenly even numbers to be a perfect square.
In fact, I believe 2^m+2^n is a perfect square if and only if either:
i) both m and n are odd and m=n, or
ii)either m or n is even (say m is even) and n=m+3.
So, the example I gave was part (i) and the example Bruno J. gave was part (ii).
Proof.
The "if" part is pretty straight forward.
So assume 2^m+2^n is a perfect square. And assume part i) doesn't hold. So we will show part ii) holds. Suppose, m $\displaystyle \leq n$.
Then 2^m+2^n=2^m(1+2^(n-m)). This implies m is even and 1+2^(n-m) is a perfect square. Write 1+2^(n-m)=k^2. Then 2^(n-m)=(k-1)(k+1). This implies k-1 is a power of 2. Write k-1=2^p. Then 2^(n-m)=2^(2p)+2^(p+1). The right hand side of the last equality can be a power of 2 only if p=1. This implies 2^(n-m)=8 or n=m+3.