I came up with $n = 4a^4 + 4a^2, \forall a \in \mathbb{Z}^+$. Then $n = 4a^4 + 4a^2 = (2a^2)^2 + (2a)^2, n + 1 = 4a^4 + 4a^2 + 1 = (2a^2+1)^2 + (0)^2, n + 2 = 4a^4 + 4a^2 + 2 = (2a^2 + 1)^2 + (1)^2$.

By Pell equation : x^2 - 2y^2 =1 has infinitely many solutions then let a=2y^2 = y^2 +y^2 we have a+1 = x^2 + 0^2 ; a+2 = x^2 + 1^2

Both of your solutions coincide with some of those given by the contestants taking part!

This was actually originally a putnam problem.

Take r odd. Saying that $a=\frac{r^2-3}2$, it works for $n=a^2$ because: $n=a^2+0^2$ $n+1=a^2+1^2$ $n+2=(a-1)^2+2a+3=(a-1)^2+r^2$

Here is mi solution: Let $n=(2x)^2=4x^2$ $\implies n+1=4x^2+1^2$ $\implies n+2=4x^2+2=(2x-1)^2+(4x+1)$ $\implies 4x+1$ it's a perfect square From here it is easy to see that $x = {6,12,20,30, ...}$ We will show that these comply: Case examples: $x=6 \implies 4x+1=25=5^2$ Case Hypothesis: We will assume that they are fulfilled until a certain $k$ Last case: $n=k+1 \implies 4(k+1)(k+2)+1=4k^2+12k+9=(2k+3)^2$ And that's it, we already show that they satisfy $x = {6,12,20,30, ...}$ with $n=4x^2$