Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 4 / 2C / iv

This is not an equivalence relation. With the usual convention that zero is a perfect square, $x^2-x^2=0$ gives reflexivity, and changing the sign of $x^2-y^2$ shows that the relation is symmetric. However $3\sim 5$ because $3^2-5^2=-16=-4^2$, and $5\sim 13$ because $5^2-13^2=-144=-12^2$, whereas $3^2-13^2=-160$ is neither a positive nor a negative perfect square: $12^2<160<13^2$. Thus transitivity fails. If “perfect square” were restricted to positive squares, reflexivity would fail as well, leaving the same negative classification.