wrong...

Remark. Essentially 2015 Turkey. Argument below is reconstructed from my solution therein. Let $m=(a+b)/2$ and $n=(a-b)/2$. Then, we have \[ \frac{m^2}{1+4(m+n)n^2} = k \in\mathbb{N}. \]Interpret this as a quadratic in $m$: \[ m^2 - \left(4kn^2\right) m -\left(4kn^3+k\right) =0. \]Call its discriminant by $4\Delta$, where \[ \Delta = 4k^2n^4 + 4kn^3+k. \]Note that $n=0$ yields the parametric family $(a,a)$; hence let $|n|>0$. Check that if $n>0$, we have \[ \left(2kn^2 +n-1\right)^2 < \Delta<\left(2kn^2+n+1\right)^2. \]From here, we must have $\Delta = \left(2kn^2+n\right)^2$, yielding $k=n^2$. The case $n<0$ is handled analogously, yields $k=n^2$. Hence, \[ m^2 = n^2\left(1+4(m+n)n^2\right). \]Now that $n\mid m$, set $m=vn$ to obtain $v^2 = 1+4n^3(v+1)$, yielding \[ v^2 - 4n^3 v-\left(4n^3+1\right)=0 \iff \left(v+1\right)\left(v-4n^3-1\right) = 0. \]As $v>0$, we get $v=4n^3+1$. This, in turn, yields $(m,n)=\left(4n^4+n,n\right)$. From here, we get $(a,b)=\left(4n^4+2n,4n^4\right)$, where $n\in\mathbb{N}$ is arbitrary.