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I want to determine whether or not $5^n+1$ , $ninmathbbN$ 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.

Say $x^2 + y^2 = 5^n + 1$.

Looking mod 2, we get that $x^2 + y^2 = 0 mod 2$, so $x$ and $y$ have the same parity, and looking mod 4, we get that $x^2 + y^2 = 2 mod 4$, so $x$ and $y$ must both be odd.

Say $x = 2r + 1$ and $y = 2s+ 1$, so $$x^2 + y^2 = (2r + 1)^2 + (2s+1)^2 = 4(r^2 + r + s^2 + s) + 2 = 5^n + 1$$

or

$$4(r(r+1) + s(s+1)) = 5^n - 1$$

Now, the left side must be divisible by $8$, so $n$ must be even.

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 : $xin mathbf N$ is a sum of two squares iff, for all primes $pequiv 3$ mod $4$, the $p$-adic valuation $v_p(x)$ is even. It's not perhaps superfluous to recall the principle of the proof : every $pequiv 1$ mod $4$ is a sum of two squares, so, by multiplicativity, it remains only to study the divisibility of $x$ by the primes $pequiv 3$ 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.

Partial observation

Let $n=2m+1$

$$5^n+1 mod 3 = 3 mod 3=0$$

So $3|5^n+1$. We can reduce the cases by observating that:
$$5^n+1equiv 0 mod 9 Rightarrow n=6k+3$$
So if $nneq6k+3$, it can't be expressed as sum of two squares because it contains $3^1$ in its factorization. In that case it may be useful:

$$5^6k+3+1=(5^2k+1+1)(5^4k+2-5^2k+1+1)$$

Improvement

Notice that :

$$5^6k+3+1equiv 0 pmod 7$$

So to be a sum of squares it must be:

$$5^6k+3+1equiv 0 pmod49 Rightarrow kequiv 3 pmod 7 $$

So our new exponents are of this form $$n_2=6(7k+3)+3=42k+21=21(2k+1)$$
Now notice that:
$$ n_1=6k+3=3(2k+1) $$
$$ n_2=42k+21=3times7(2k+1) $$

If we could prove that this process goes on infinitely than the thesis would be demonstrated. But i don't know how.

:/