Is any power of four equal to the sum of two different squares?

Question

More simply, is there a solution for:

$x^2+y^2=4^n$

where x, y, and n are positive integers and $x\neq y$.

My gut feeling says no, but I can't justify it with an actual proof.

If the answer is no, is there a simple proof as to why?

Answer 1

Assume that $x,y$ are solutions to the equation $x^2+y^2=4^n$. Then we can write $x=2^ka,y=2^\ell b$ with $a,b$ odd, so we get $4^ka^2+4^\ell b^2=4^n$. WLOG $k\leq \ell$ so we have $a^2+4^{\ell-k}b^2=4^{n-k}$. If $k\neq\ell$, then we immediately have a contradiction because the left-hand side is odd and the right-hand side is even. Thus $k=\ell$. Therefore we have that $a^2+b^2=4^{n-k}=4^{n-\ell}$ where $a,b$ are odd. The square of any odd number is $1\pmod{4}$, so the left-hand side is $2\pmod{4}$. But the right-hand side is never $2\pmod{4}$ so there are no solutions.

Note: This doesn't use the assumption in the OP that $x\neq y$

Answer 2

NO. (1). If $x,y$ are both odd then $x^2+y^2$ is congruent to $2$ modulo $4.$

(2). If there were a least $n$ such that $x^2+y^2=4^n$ with $x\ne y$, then, as $x,y$ must both be even, let $x=2x'$ and $y=2y'$.

Then $0\ne x'\ne y'\ne y$ so $1<x'+y'\leq x'^2+y'^2=4^{n-1},$ so $n-1\geq 1.$ So $4^{n-1}$ is also a sum of two non-zero unequal squares, contrary to the minimality of $n.$

Another way of saying this is that if S is the set of all $n\in \mathbb N$ such that $4^n$ is the sum of $2$ unequal non-zero squares, then $n\in S\implies (n-1)\in S.$ A set $S\subset \mathbb N$ with the property that $n\in S\implies (n-1)\in S$ must be the empty set. (Else it would be a non-empty set of natural numbers with no least member.)