Find the number of ordered pairs $$(𝑚, 𝑛)$$ where $$𝑚$$ and $$𝑛$$ are positive integers such that $$1 ≤ 𝑚 < 𝑛 ≤ 50$$ and the product $$𝑚𝑛$$ is a perfect square.

Write each positive integer in the form $$k\;x^2$$ where
  • $$k$$ is square-free (no prime square divides it),
  • $$x$$ is a positive integer.

If $$m=k\,u^2$$ and $$n=k\,v^2$$ then
$$mn=(k\,u^2)(k\,v^2)=k^2(u\,v)^2=(k\,u\,v)^2,$$
which is a perfect square. Conversely, if $$mn$$ is a perfect square, the parity of every prime’s exponent in $$m$$ and $$n$$ must be the same, so their square-free parts are identical. Hence

mn is a perfect square $$\Longleftrightarrow$$ $$m=k\,u^2,\;n=k\,v^2$$ with the same square-free $$k$$.

Thus for every square-free $$k\le 50$$ count ordered pairs $$(u,v)$$ with $$u\lt v,\qquad k\,u^2\le 50,\;k\,v^2\le 50.$$ Put

$$t(k)=\Bigl\lfloor\sqrt{\frac{50}{k}}\Bigr\rfloor$$ (the number of positive integers $$x$$ satisfying $$k\,x^2\le 50$$). For a fixed $$k$$ the choices of $$(u,v)$$ are all $$\binom{t(k)}{2}$$ pairs with $$u\lt v$$.

List every square-free $$k\le 50$$, evaluate $$t(k)$$, keep only those with $$t(k)\ge 2$$:

$$\begin{array}{c|c|c} k & t(k)=\left\lfloor\sqrt{50/k}\right\rfloor & \binom{t(k)}{2} \\ \hline 1 & 7 & 21\\ 2 & 5 & 10\\ 3 & 4 & 6\\ 5 & 3 & 3\\ 6 & 2 & 1\\ 7 & 2 & 1\\ 10 & 2 & 1\\ 11 & 2 & 1 \end{array}$$

All other square-free $$k$$ (13,14,15,17,19,\,$$\dot$$s,47) give $$t(k)=1$$, hence contribute 0 pairs.

Add the contributions:

$$21+10+6+3+1+1+1+1=44.$$ Therefore the number of ordered pairs $$(m,n)$$ with $$1\le m\lt n\le 50$$ for which $$mn$$ is a perfect square equals

44.