= Solution
\b[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 $3\sim5$ because $3^2-5^2=-16=-4^2$, and $5\sim13$ 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.
Back to article page
