A relation in a finite ring

Question

Let $R$ be a finite ring such that for any $a,b\in R$ there exists $c\in R$ (depending on $a$ and $b$) such that $a^2+b^2=c^2$.

Prove that for any $a,b,c\in R$ there exists $d\in R$ such that $2abc=d^2$.

Answer 1

Use the word "assumption" to refer to the statement: "for any $a, b\in R$ there exists $c\in R$ (depending on $a$ and $b$) such that $a^2+b^2=c^2$". Let the phrase ``desired conclusion'' refer to the statement: "for any $a, b, c\in R$ there exists $d\in R$ such that $2abc=d^2$".

The assumption implies that the set of squares is closed under sum. This implies that if $u$ is a square, then $u+u=2u$ is a square, and $u+2u=3u$ is a square, etc, until $-u$ is a square (by finiteness). It follows from this that $(x+1)^2 + (-x^2) + (-1) = 2x$ is a square for any $x$, which is the desired conclusion when $x=abc$.