TITLE: Prove that $2^n+3^n $ is never a perfect square
QUESTION [11 upvotes]: My attempt :
If $n$ is odd, then the square must be 2  (mod 3), which is not possible.
Hence $n =2m$
$2^{2m}+3^{2m}=(2^m+a)^2$
$a^2+2^{m+1}a=3^{2m}$
$a (a+2^{m+1})=3^{2m} $
By fundamental theorem of arithmetic, 
$a=3^x $
$3^x +2^{m+1}=3^y $
$2^{m+1}=3^x (3^{y-x}-1) $
Which is not possible by Fundamental theorem of Arithmetic 
Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.

REPLY [2 votes]: Yet another approach. 
For the odd case of $n$, as you mentioned the result is  $2\mod 3$ which cannot be a square number.
As for the even case $n=2m$, we have $2^{2m}+3^{2m}\mod 10$ being equal to $2$ (for when $m=0$) or either $3$ (for $n=2+4k,k\geq0$) or $7$ (for $n=4k,k\geq1$), which clearly cannot be square (as a square number modulus $10$ takes values $0,1,4,9,6,5$).