For the list of relevant papers see
>>> this page (click) <<<
Challenge: Prove the following
If a quadruple of four non-negative integers $A$, $B$, $C$, $D$ is primitive, i.e., $\gcd(A,B,C,D)=1$, satisfies $$2(A^2 + B^2 + C^2 + D^2) = (A+B+C+D)^2$$
then any two add up to a sum of two squares:
$$A+B = p^2+q^2$$
for some integers $p$ and $q$.
No comments:
Post a Comment