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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?

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    $\begingroup$ 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 $z^2-2y^2=-1$. $\endgroup$ – Galc127 Dec 16 '18 at 14:02
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    $\begingroup$ The list of values for $y$ is OEIS A$001653$. That link has multiple descriptions, including connections to Pell's Equation. $\endgroup$ – lulu Dec 16 '18 at 14:03
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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:

$$2(y)^2-[2x+1]^2=1=2(3y+4x+2)^2-[4y+6x+3]^2=1 \\ \implies \text{The recursive relation:} $$

$$\begin{cases} y_{k+1}=3y_k+4x_k+2 \\ x_{k+1}=2y_k+3x_k+1 \end{cases}$$

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