# There exist infinitely many pairs of consecutive squares s.t. their sum is also a square

First of all, the term "pairs" is two of them, I assume (question's formulation is rather difficult to understand for me). So I guess this is the statement: $$\text{There exist infinitely many pairs of consecutive squares }x^2,\ (x+1)^2\text{ s.t. }x^2+(x+1)^2=y^2\ x,y\in\mathbb{N}.$$

So there are two theorems I know of that might be of any use here:

1. An odd prime $$p$$ can be written as sum of squares iff $$p\equiv1$$ mod $$4$$.

2. $$n\in\mathbb{N}$$ is the sum of squares iff primes that are $$3$$ mod $$4$$ occur an even number of times in the prime factorisation of $$n$$.

We can work out the LHS like so: $$x^2+(x+1)^2=2x(x+1)+1$$. We know that either $$x$$ or $$x+1$$ must be even; hence, the expression is of the form $$4k+1$$ with $$k\in\mathbb{Z}$$. From this we can conclude that the sum of two consecutive squares is always congruent $$1$$ modulo $$4$$ which also implies that primes that are $$p\equiv3$$ mod $$4$$ that divide the sum of the consecutive squares must occur an even number of times. To this point, this does not prove anything significant, I think.

Does anyone have a hint on how to prove the statement?

• You want $$2x^2+2x+1=y^2\iff 4x^2+4x+1+1=2y^2\iff (2x+1)^2+1=2y^2$$ and this is a Pell equation of the form $z^2-2y^2=-1$. – Galc127 Dec 16 '18 at 14:02
• The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation. – lulu Dec 16 '18 at 14:03

This question was answered here, and in Galc127's comment above. Basically, the above relation is the pell equation $$2y^2-(2x+1)^2=1$$, and the cool thing about pell equations is that when you have one solution, you have a countable infinity of them, visibly demonstrated by the identity below:
$$2(y)^2-[2x+1]^2=1=2(3y+4x+2)^2-[4y+6x+3]^2=1 \\ \implies \text{The recursive relation:}$$
$$\begin{cases} y_{k+1}=3y_k+4x_k+2 \\ x_{k+1}=2y_k+3x_k+1 \end{cases}$$