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?

  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ 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

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:

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.