Infinitely many $n$ such that $n, n+1, n+2$ are each the sum of two perfect squares.

n = 0 is a trivial solution as 0 = $0^2 + 0^2, 1 = 1^2 + 0^2, 2 = 1^2 + 1^2$. Consider the pell equation $x^2 - 2y^2 = 1$, which has infinitely many solutions. Take $n = x^2 -1$. This implies $ n = y^2 + y^2$, $n+1 = x^2 + 0^2$ and $n+2 = x^2 + 1^2$.

One such infinite family is $$(2a^2 + 1)^2 - 1, (2a^2 + 1)^2 + 0^2, (2a^2 + 1)^2 + 1^2$$ for all integer $a$. This works because $$(2a^2 + 1)^2 - 1 = 4a^4 + 4a^2 = (2a^2)^2 + (2a)^2.$$

If we can find $y$ such that $y^2-1$ is a sum of two squares, say $y^2- 1 = a^2+b^2$, then take $n = y^2-1$ and we'll have $n = y^2-1 = a^2+b^2$, $n+1 = y^2-1 +1 = y^2 +0$, $n+2 = y^2 + 1$. So, we want to show the existence of infinitely many $y$ such that $y^2-1$ is a sum of two squares.

Look at $3^{2^k}$ for integers $k \geq 1$. For $k = 1$, $3^2 - 1= 8 = 2^2 + 2^2$.

Now, $3^{2^2} -1 = 3^4 -1 = (3^2-1)(3^2+1) = (2^2 +2^2)(3^2+1^2) $

Note the identity $(p^2+q^2)(c^2+d^2) = (pc+qd)^2 + (pd-qc)^2$. So we know that the product of two sums of two squares is again a sum of two squares. And hence, without explicitly calculating, we can say $3^{2^2}-1$ is a sum of two squares.

Now suppose for the sake of induction that $3^{2^k} - 1$ is a sum of two squares. Then $3^{2^{k+1}} -1 = (3^{2^k}-1)((3^{2^{k-1}})^2 +1^2)$. Thus from the induction hypothesis and our neat identity, $3^{2^{k+1}} -1$ is a sum of two squares.

So, for all integers $k \geq 1$, taking $y = 3^{2^k}$ and $n = y^2-1$gives us infinitely many integers satisfying the problem's requirement.