WakeUp wrote: Determine all integers $n$ for which \[\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}\] is an integer. In order expression be defined, we need $n\in[0,156]$ Let $m=\sqrt{\frac{25}2+\sqrt{\frac{625}4-n}}$ $+\sqrt{\frac{25}2-\sqrt{\frac{625}4-n}}$ $\iff$ (squaring) $m>0$ and $\sqrt n=\frac{m^2-25}2$ $\iff$ (squaring) $m^2\ge 25$ and $n=\left(\frac{m^2-25}2\right)^2$ So we need $m$ odd $\in[5,7]$ (upper bound for $n\le 156$) Hence the two solutions $\boxed{n\in\{0,144\}}$

pco wrote: WakeUp wrote: Determine all integers $n$ for which \[\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}\]is an integer. In order expression be defined, we need $n\in[0,156]$ Let $m=\sqrt{\frac{25}2+\sqrt{\frac{625}4-n}}$ $+\sqrt{\frac{25}2-\sqrt{\frac{625}4-n}}$ $\iff$ (squaring) $m>0$ and $\sqrt n=\frac{m^2-25}2$ $\iff$ (squaring) $m^2\ge 25$ and $n=\left(\frac{m^2-25}2\right)^2$ So we need $m$ odd $\in[5,7]$ (upper bound for $n\le 156$) Hence the two solutions $\boxed{n\in\{0,144\}}$ This solution is incorrect; take for example n = 28^2, then the expression evaluates to 9. $n$ can take on any element in the sequence $\left(\frac{m^2 - 25}{2}\right)^2$, for $m$ any odd perfect square at least 25.

Khazix wrote: pco wrote: [...] This solution is incorrect; take for example n = 28^2, then the expression evaluates to 9. $n$ can take on any element in the sequence $\left(\frac{m^2 - 25}{2}\right)^2$, for $m$ any odd perfect square at least 25. pco's solution isn't incorrect (in fact I don't think I've ever seen pco post a solution that's incorrect). The only discrepancy between your solution and his is the issue on whether $\sqrt{\dfrac{625}{4}-n}$ is allowed to be imaginary, which is up for debate; however, the essence of both of your solutions are the same.