when could $5^m+5^n$ represented as sum of two squares

Question

How can we prove that $5^m+5^n$ could be expressed as a sum of two squares if and only if $m-n$ is even with $m,n\in\mathbb{Z}_{>0}$

I was able to prove that any power of $5$ could be expressed as a sum of two squares but I can't prove the required result, can someone help me please?

Answer 1

It's quite trivial without using any kind of SST or such theorems:

1. If $m\overset{2}{\equiv}n\overset{2}{\equiv}0$ then we're done since $5^m+5^n=\left(5^{\frac m2}\right)^2+\left(5^{\frac n2}\right)^2$
2. Otherwise if $m\overset{2}{\equiv}n\overset{2}{\equiv}1$ then since $m-1\overset{2}{\equiv}n-1\overset{2}{\equiv}0$ one could use the result of the former case and write: $$5^m+5^n=5(5^{m-1}+5^{n-1})=(1+4)\left(\left(5^{\frac{m-1}2}\right)^2+\left(5^{\frac {n-1}2}\right)^2\right)=\left(\left(5^{\frac{m-1}2}\right)+2\left(5^{\frac {n-1}2}\right)\right)^2+\left(2\left(5^{\frac{m-1}2}\right)-\left(5^{\frac{n-1}2}\right)\right)^2$$ Finally for the other side of the theorem note that if $m,n$ are of different parities, $5^m+5^n\overset{8}{\equiv}6$ which can't be sum of two perfect squares as perfect squares generate the set of residues $0,1,4 \mod8$

Answer 2

Let $m-n$ be even and $m>n$. Then $5^m+5^n=5^n(5^{m-n}+1)$. $5^n$ is a sum of two squares as you stated, $5^{m-n}+1$ is also a sum of two squares. So due to the https://en.wikipedia.org/wiki/Brahmagupta–Fibonacci_identity $5^m+5^n$ is also a sum of two squares.

Now let $m-n$ be odd, so $m$ and $n$ are of different parity. Look at $5^m+5^n$ modulo $8$. One of the summands is $5$ while the other is $1$. So $5^m+5^n$ is $6$ modulo $8$. But a sum of two squares is never $6$ modulo $8$ because squares can only be $0,1$, or $4$ modulo $8$.