WakeUp wrote: Find all $n\in\mathbb{Z}$ such that the number $\sqrt{\frac{4n-2}{n+5}}$ is rational. So $4n-2=ap^2$ and $n+5=aq^2$ and so $4aq^2-22=ap^2$ and so $a(2q-p)(2q+p)=22$ Since $4\not | 22$, $2q-p$ and $2q+p$ both are odd and $a=\pm 2$ which gives eight possibilities : $2q-p=1$ and $2q+p=11$ $\implies$ $p=5$ and $q=3$ and $a=2$ and $n=13$ $2q-p=1$ and $2q+p=-11$ $\implies$ no solution $2q-p=-1$ and $2q+p=11$ $\implies$ no solution $2q-p=-1$ and $2q+p=-11$ $\implies$ $p=-5$ and $q=-3$ and $a=2$ and $n=13$ $2q-p=11$ and $2q+p=1$ $\implies$ $p=-5$ and $q=3$ and $a=2$ and $n=13$ $2q-p=11$ and $2q+p=-1$ $\implies$ no solution $2q-p=-11$ and $2q+p=1$ $\implies$ no solution $2q-p=-11$ and $2q+p=-1$ $\implies$ $p=5$ and $q=-3$ and $a=2$ and $n=13$ Hence the unique solution : $\boxed{n=13}$

equivalent saying there is some x,y and k s.t. $4n-2=kx^2 $ and $n+5=ky^2$ but then $k(4y^2-x^2)=22$ so, k can be 1,2,11,22 if k=1 then $(4y^2-x^2)=22$ which has no integral solutions. if k=2 the $(4y^2-x^2)=11$ which has only solution y=3 and x=5 if k=11 then $(4y^2-x^2)=2$ which has no integral solutions. if k=22 then $(4y^2-x^2)=1$ which has no integral solutions. so only possibility= $n=ky^2+5=2*3^2-5=13$ see it is indeed a solution when $n=13$ $ \sqrt{\frac{4n-2}{n+5}}=\sqrt\frac{50}{9}=\frac{5}{3}$