If I'm correct, it even suffices to know that $ab$ divides $a^2+b^2-a$ (without the factor of $2$)...

Tintarn wrote: But clearly also $r^2s^2 \mid a^2+b^2$ How did you get that...?

By setting $d = GCD(a,b)$, we obtain $(a,b) = (sd,td)$, where $s$ and $t$ are two positive coprime integers. Consequently $(1) \;\; \frac{a^2 + b^2 - a}{2ab} = \frac{(sd)^2 + (td)^2 - sd}{2 \cdot (sd) \cdot (td)} = \frac{d(s^2 + t^2) - s}{2dst} \in \mathbb{N}$. Therefore $s | dt^2$ by (1), yielding $s | d$ since $GCD(s,t)=1$. Futhermore $d | s$ by (1), which means $d=s$ (since $s | d$). Hence $a = sd = s \cdot s = s^2$, i.e. $a$ is a perfect square. q.e.d.

duck_master wrote: Tintarn wrote: But clearly also $r^2s^2 \mid a^2+b^2$ How did you get that...? Well, $rs \mid a$ and $rs\mid b$ so $r^2s^2 \mid a^2$ and $r^2s^2 \mid b^2$ so $r^2s^2 \mid a^2+b^2$...

Can anyone tell whether this solution is correct or not.

Kindly if anyone can tell whether my solution was correct or not.

Similar to BWM 2013 P5 BWM 2013 P5 wrote: Let $m,n \in \mathbb{Z}^+$, such, $$mn | m^2+n^2+m$$Prove, $m$ is a perfect square Bosnia-Herzegovina Regional Olympiad 2016 Grade 10 P2 wrote: Let $a$ and $b$ be two positive integers such that $2ab$ divides $a^2+b^2-a$. Prove that $a$ is perfect square Solution: Let $\gcd (a,b)=d$ $\implies$ $\begin{cases} a=md \\\ b=nd \\\ \gcd (m,n)=1 \end{cases}$ $$\frac{a^2+b^2-a}{2ab} =\frac{m^2d^2+n^2d^2-md}{2mnd^2} =\frac{m^2d+n^2d-m}{2mnd} \in \mathbb{Z}$$Since, $m$ divides $2mnd$ $\implies$ $m ~ | ~ m^2d+n^2d-m $ but since, $\gcd (m,n)=1$ $\implies$ $m ~ | ~ d$ And, Since, $d$ divides $2mnd$ $\implies$ $d ~ | ~ m^2d$ $+n^2d-$ $m$ $\implies$ $d ~ | ~ m$ $$\implies m=d \implies a=md=m^2=d^2$$Hence done! $\qquad \blacksquare$