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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 $2ab = c^2 - 1$. -

Mirko
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Mike
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  • I think you can use Vietta's method. – Xam Nov 03 '16 at 23:35
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    Always better to give a title that gives some idea of what the question is. – Thomas Andrews Nov 04 '16 at 00:24
  • of $m$ and $n$ exactly one is odd, and the one that is even is divisible by $8$. Use that $(n^2+m^2+1)/2$ is odd (use that odd$^2-1=0\mod4$). The recursive sequence $M_0=0,M_1=1,M_{k}=2M_{k-1}+M_{k-2}$, produces $0,1,2,5,12,29,70,169,408,985,2378,5741,13860$,etc, nonzero consecutive values give $n,m$ for which the fraction $\displaystyle\frac{(n^2+m^2+1)(n^2+m^2-1)}{2mn(n+m)(n-m)} = \pm2$. Googled it: https://oeis.org/A000129 Pell numbers, sometimes also called lambda numbers – Mirko Nov 04 '16 at 04:44
  • @Mirko are you claiming that if $2mn(n+m)(n-m)$ divides $(n^2 + m^2 + 1)(n^2 + m^2 - 1)$ with $n,m > 0$ then the pair $n,m$ are consecutive Pell-numbers? - which, by the way, seems very likely if you run program and test the claim. – Mike Nov 04 '16 at 11:44
  • it seems that if $m,n$ are consecutive Pell-numbers then $2mn(n+m)(n-m)$ divides $(n^2+m^2+1)(n^2+m^2-1)$ and the quotient is $2$. It might not be too difficult to prove this part, thought I didn't try.(The Pell numbers are denominators in $\sqrt2$ approximations.) It is (your linked) question, whether the quotient is $2$ whenever $2mn(n+m)(n-m)|(n^2+m^2+1)(n^2+m^2-1)$, and whether, in addition, $m,n$ must be consecutive Pell-numbers. I cannot rule out that either the quotient is $2$ but $m,n$ need not be consecutive Pell-numbers, or that the quotient might be an integer different from $2$. – Mirko Nov 04 '16 at 15:55
  • editing/typos/comment: Please consider stating this as a question, e.g. "then is it true that $n,m$ are a pair of..,etc. Also,in the notice, change the second $m<n$ (that goes with $−2$) to $n<m$. Well,ok I just did this edit, please take a look to make sure it was ok. I ran a program that gave a sequence of numbers between $2$ and $13860$. When I googled those numbers, I realized they were Pell's numbers. The program tested whether $2mn(n+m)(n-m)$ divided $(n^2+m^2+1)(n^2+m^2-1)$ and only produced pairs of consecutive Pell numbers. So perhaps there are no other solutions, but I do not know. – Mirko Nov 04 '16 at 17:15
  • I think it was pretty clear before your edit, but thanks anyway. Well I have shown that if $n,m$ are consecutive Pell-numbers, then the statement does hold see: http://math.stackexchange.com/questions/2000382/an-identity-for-pell-numbers/2000605#2000605 – Mike Nov 05 '16 at 19:43
  • But I am really quite lost in trying to show that these are the only solutions! If you have any suggestions on this, they would be much appreciated. – Mike Nov 05 '16 at 19:44
  • computer algebra shows the solutions of $Q(n,m):=\frac{(n^2+m^2)^2-1}{2mn(n^2-m^2)}=2$ are $n=\pm\sqrt{2m^2\pm1}+m$ so Pell's equation is involved when the ratio is $2$ (to show only solutions). Something similar could be attempted for $Q(n,m)=k$ (computer is lost, but this is 4th degree equation for $n$ in $m,k$). The function $Q(n,m)$ for real $n>m\ge1$ seems to have minimum(s) (when either $n$ or $m$ is fixed) slightly less than $2$ achieved at(or near?) $n=(1+\sqrt{2})m$. If $h(m)=Q((1+\sqrt{2})m,m)$ then $h(1)≈1.957$, $h(3)≈1.9997$, $h(30)≈1.99999995$. Look at an Excel table for $Q(n,m)$ – Mirko Nov 06 '16 at 03:56

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