Pythagorean triples and Pell numbers.

This problem is mentioned in this one, but I think it deserves some attention on its own. So here it is:

For any integers $n,m > 0$:

If $2mn(n+m)(n-m)$ divides $(n^2 + m^2 + 1)(n^2 + m^2 – 1)$, then is it true that $n,m$ are a pair of consecutive Pell-numbers, where the Pell-numbers are given recursively by:

$P_0 = 0$

$P_1 = 1$

$P_{n+2} = 2P_{n-1} + P_{n-2}$.

The converse is definitely true.

See Wikipedia article on Pell numbers, and the OEIS page.

Remark:

Notice that this implies the quotient of $(n^2 + m^2 + 1)(n^2 + m^2 – 1)$ by
$2mn(n+m)(n-m)$ is then 2 if $m < n$ and $-2$ if $n < m$. So that in the case when $n,m$ are coprime we have following, if the above is true:

Let $(a,b,c)$ be a primitive Pythagorean triple. If $ab \mid c^2 – 1$, then $2ab = c^2 – 1$. –

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