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

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  • $\begingroup$ I think you can use Vietta's method. $\endgroup$ – Xam Nov 3 '16 at 23:35
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    $\begingroup$ Always better to give a title that gives some idea of what the question is. $\endgroup$ – Thomas Andrews Nov 4 '16 at 0:24
  • $\begingroup$ 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: oeis.org/A000129 Pell numbers, sometimes also called lambda numbers $\endgroup$ – Mirko Nov 4 '16 at 4:44
  • $\begingroup$ @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. $\endgroup$ – Mike Nov 4 '16 at 11:44
  • $\begingroup$ 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 $2$. $\endgroup$ – Mirko Nov 4 '16 at 15:55

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