Can $5^n+1$ be sum of two squares?

Question

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.

Answer 1

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: $$5^n+1\equiv 0 \mod 9 \Rightarrow n=6k+3$$ So if $n\neq6k+3$, 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 :

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

So to be a sum of squares it must be:

$$5^{6k+3}+1\equiv 0 \pmod{49} \Rightarrow k\equiv 3 \pmod 7 $$

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

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

:/

Answer 2

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.

Answer 3

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.