For completeness, this problem was A2 at the contest. We are looking for solutions with $ n = 4x^2$. Then $ n = (2x)^2 + 0^2$, $ n + 1 = (2x)^2 + 1^2$. It remains to find infinitely many $ x$ for which there exist $ y,z\in\mathbb{Z}$ such that $ n + 2 = 4x^2 + 2 = y^2 + z^2$, for some $ y,z$. We'll search for solutions having $ y = z$, that is $ 2x^2 + 1 = y^2$. This is a simple Pell Equation and it has infinitely many solutions, all of the form $ (y,x) = \left(\dfrac{(3 + 2\sqrt2)^n + (3 - 2\sqrt2)^n}2,\dfrac{(3 + 2\sqrt2)^n - (3 - 2\sqrt2)^n}{2\sqrt2}\right)$. It is easy to see that the above numbers are integers and indeed satisfy $ y^2 - 2x^2 = 1$.

another (somehow similar) solution is the following: we know that th equation $ k^2-2t^2=1$ has infinite many solutions,now put: $ n=2t^2\rightarrow n=t^2+t^2$ $ \Rightarrow n+1=k^2\rightarrow n+1=k^2+0^2$ $ \Rightarrow n+2=k^2+1\rightarrow n+2=k^2+1^2$

Another solution: Consider the following sequence: $ a_{1}=8$, $ a_{n+1}=a_{n}(a_{n}+2)$ We claim that for each $ n$, $ (a_{n}, a_{n}+1, a_{n}+2)$ are three consecutive integers, each being the sum of two squares. We will prove that by induction. For $ n=1$, we have $ 8=4+4, 9=9+0, 10=9+1$. Now assume that $ (a_{n}, a_{n}+1, a_{n}+2)$ are sum of two squares. Then we have $ a_{n+1}=a_{n}(a_{n}+2) \longrightarrow$ being the sum of two squares by Lagrange's lemma. $ a_{n+1}+1=a_{n}(a_{n}+2)+1=(a_{n}+1)^2+0$ $ a_{n+1}+2=a_{n}(a_{n}+2)+2=(a_{n}+1)^2+1$ The proof is complete.