Can $5^n+1$ be sum of two squares? I want to determine whether or not $5^n+1$ , $n\in\mathbb{N}$ can be written as sum of two squares.
Obviously, the real problem is when $n$ is odd. I am aware of the known results about numbers written as sum if squares, but I couldn't apply them here. We can see that when $n$ is odd, $5^n+1$ ia divisible by 6 and gives reminder 2 when divided by 4 which means that the two squares we would have to add must end in 1 and 9 or 5 and 5.

You say that you're "aware of the known results about numbers written as sums of squares". Let's try to apply them here (just for fun, because the solution by @Michael Biro is perfectly elementary). We have the classical theorem : $x\in \mathbf N$ is a sum of two squares iff, for all primes $p\equiv 3$ mod $4$, the $p$-adic valuation $v_p(x)$ is even. It's not perhaps superfluous to recall the principle of the proof : every $p\equiv 1$ mod $4$ is a sum of two squares, so, by multiplicativity, it remains only to study the divisibility of $x$ by the primes $p\equiv 3$ mod $4$. Such a prime $p$ is inert (i.e. remains prime) in the PID $\mathbf Z[i]$, so if $x=a^2+b^2=(a+bi)(a-bi), p$ divides both the factors.
Now $3$ is inert and $5$ splits as $(2+i)(2-i)$ in $\mathbf Z[i]$. But $ 5^{2m+1} = (6-1)^{2m+1}=6^{2m+1}- 6^{2m}+...+6-1$ by the binomial formula, so $ 5^{2m+1}+1=6(6^{2m}-...+1)$, whose $3$-adic valuation is $1$. This shows that the exponent in your problem must be even. 
