sqing wrote: Find all the pairs of integers $ (m, n)$ such that $ \sqrt {n +\sqrt {2016}} +\sqrt {m-\sqrt {2016}} \in \mathbb {Q}.$ First of all it is obvious that $m\geq 45$. Let us assume that $m\neq n$. Then there exists $a\in\mathbb{Q}$ such that $$ \sqrt {n +\sqrt {2016}} +\sqrt {m-\sqrt {2016}}=a \Leftrightarrow \displaystyle{\sqrt{2016}=\displaystyle\frac{(\frac{a^2-n-m}{4})^2-nm}{m-n}}$$But the LHS is irrational, whereas the RHS is rational, contradiction. So $m=n$ and the equation becomes $$2\sqrt{n^2-2016}=a^2-2n $$with $a^2\ge 2n$. Hence, $n^2-2016=q^2\;\;(1)$, where $q\in\mathbb{Q}$. But $n^2-2016\in\mathbb{Z}$,so $q\in\mathbb{Z}$. Therefore $a\in\mathbb{Z}$. From (1) we have $$(n-q)(n+q)=2016 $$So $n=505,254,171,130,90,79,71,65,54,50,46,45$ We easily see that only $n=50,65,130,505$ satisfy the initial inequality.

$\sqrt {n +\sqrt {2016}} +\sqrt {m-\sqrt {2016}}=a \in \mathbb {Q}$ $m+n+2\sqrt{(n+\sqrt{2016})(m-\sqrt{2016})} =a^2 \in \mathbb {Q}$ $(n+\sqrt{2016})(m-\sqrt{2016})\in \mathbb {Q}$ $mn-2016+(m-n)\sqrt{2016} \in \mathbb {Q} \to m=n$ $2m+2\sqrt{m^2-2016} =a^2 \in \mathbb {Q}$ $m^2=2016+t^2 \to t \in \mathbb{N}$ $2m+2t = a^2 \in \mathbb{N}$ $a^2>2m>2\sqrt{2016}>80$ $2*2016 = 2(m^2-t^2)=a^2(m-t)$ $a^2|2*2016=2^6*3^2*7$ $a^2= 144,576$ Case 1: $a^2=144 \to m-t=28,m+t=72 \to m=50$ Case 2:$a^2=576 \to m-t=7, m+t=288 \to $ contradiction. So $m=n=50$ and $a=12$

Where can we find JBMO shortlist? Or is it unrelased?

Same question, where can we find JBMO shortlist?

Duarti wrote: Same question, where can we find JBMO shortlist? The most efficient way is trying to get it from your country's leader of that JBMO. He/She certainly has it.

U posted some nice solutions, but wouldn't it be easier if you went old fashion way when solving integer equality, just translating one of the numbers form left to right then squaring up? I mean, I managed to get m = n this way in somewhat less lines. Just asking to see if I maybe made a mistake or two.