Gazeta Matematica 12/2015, Problem S:E15.354

Problem:
Prove that the number $$a=2n^2+[\sqrt{4n^2+n}]+1$$ where $n$ is a nonzero natural number, can be written as the sum of two perfect squares.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let us prove first that $[\sqrt{4n^2+n}]=2n$. Indeed, for all $n \in \mathbb{N}$ $$4n^2 \leq 4n^2+n < 4n^2+4n+1,$$ so $$2n \leq \sqrt{4n^2+n}<2n+1.$$ It follows that $[\sqrt{4n^2+n}]=2n$. So, $$a=2n^2+2n+1=n^2+(n+1)^2.$$