TITLE: Infinitely many $n$ such that $n, n+1, n+2$ are each the sum of two perfect squares.
QUESTION [8 upvotes]: Prove that there exist infinitely many integers $n$ such that $n$, $n+1$, $n+2$ are all the sum of two perfect squares. Induction does not seem to be yielding any results.

REPLY [3 votes]: I think it's better to solve the problem for the General case. For the case of even numbers the solution there.  Mohanty's conjecture
For odd numbers.
$$\left\{\begin{aligned}&N=a^2+b^2\\&N+T=c^2+q^2\\&N+2T=x^2+y^2\end{aligned}\right.$$
Decompose the number into factors.  $T=ps$
The solution can be written as.
$$a=pk^2+(2p+s)k+\frac{p+3s}{2}$$
$$b=pk^2+(2p+s)k+\frac{3p+s}{2}$$
$$c=pk^2+(3p+s)k+\frac{3(p+s)}{2}$$
$$q=pk^2+(p+s)k-\frac{p-s}{2}$$
$$x=pk^2+(2p+s)k+\frac{p+s}{2}$$
$$y=pk^2+(2p+s)k+\frac{3(p+s)}{2}$$
It is seen that for all odd numbers are infinitely many solutions, not just $1$.