Question 26

Every positive integer $n$ can be written uniquely as $n = s(n) \cdot k^2$, where $s(n)$ is square-free. Then $xy$ is a perfect square if and only if $s(x) = s(y)$:

• Reflexive (A): For any $x$, $x \cdot x = x^2$ is a square, so $(x, x) \in R$.
• Symmetric (B): If $xy = m^2$, then $yx = m^2$, so $(y, x) \in R$.
• Transitive (C): If $xy = m^2$ and $yz = k^2$ (with $y > 0$), then $xz = \frac{m^2 k^2}{y^2} = \left(\frac{mk}{y}\right)^2$, which is an integer square because $s(x) = s(y) = s(z)$.
• Not Antisymmetric (D): $(1, 4) \in R$ and $(4, 1) \in R$ since $1 \times 4 = 2^2$, but $1 \neq 4$.