This question originally from Gauss's Circle Problem. I want to prove that there are exactly 4 integer points on the circle $$x^2+y^2=2^n$$ Where $n\in \mathbb N$, and $x,y \in \mathbb Z$
I split this question into two cases.
Prove there are only 4 integer roots on circle $x^2+y^2=2^{2n+1}$
And Prove there are only 4 integer roots on circle $x^2+y^2=2^{2n}$
From the observation, all the even power cases will include 1 zero in the roots.
If the statement is true, all the odd power cases will have 4 roots by contradiction.
It is obvious that $(2^{n})^2+(2^{n})^2=2^{2n+1}$
Because of $x,y$ can be negative integers, we find four roots for this circle.
So we need to prove that there is no more integers satisfied the equation.
Assume there are other integers pairs $(a,b)$ can satisfy this equation.
Without losing generality, I assume that $a\gt 2^n\gt b$
Thus $(a^2+b^2)(2^n+2^n)=2^{4n+2}=2^{2(2n+1)}=(a2^n+b2^n)^2+(a2^n-b2^n)^2$
If one of the solutions is zero, then it must be $(a2^n-b2^n)$
Thus $a=b$
Which contracts with my assumption.
So how could you prove that there is must be one zero in the even power cases?
If $x^2 + y^2 \equiv 0 \pmod 4,$ then both $x,y$ are even, we can divide them by $2$ and the target number by $4.$ If the original number is a power of $4,$ we descend to $x^2 + y^2 = 1,$ which has four solutions $(1,0); (-1,0); (0,1); (0,-1)$