# For m,n integers, prove (or disprove) that if $n(n-1) | (m-1)(m+1)$ then

either $n | m-1$ or $n | m+1$ or $n-1|m-1$ or $n-1|m+1$. Tried quite a bit but no luck yet. Thanks

EDIT: Sorry but I want to state that none of n, n-1, m-1, m+1 is prime.

• This seems like something that should only be true for cases where some or all of those numbers are prime. Have you tried any examples? Apr 12, 2017 at 22:44
• yes... tried examples. I feel it is true for non-prime m,n. But looking for proof.
– sku
Apr 12, 2017 at 22:48
• With problems like these, if I don't see an intuitive lead why it should be true, I'm usually quick to switch gears and write a python script. If no counter examples lie within immediate reach, then I'll go back to head scratching. In this case, counter examples lie just beyond pen and paper territory. Apr 12, 2017 at 23:22

It is not true, here's the minimal counterexample I could come up with:

$n=21, m=29$,

Then $n(n-1) = 21 \cdot 20 = 420 | 840 = 28 \cdot 30 = (m-1)(m+1)$

and you can easily see that none of the two factors on the right is a multiple of one on the left.

Counter example $n=16$ & $m=71$.

$(n,m-1)=(16,70)$ does not divide. $(n,m+1)=(16,72)$ does not divide.$(n-1,m+1)=(15,70)$ does not divide.$(n-1,m+1)=(15,72)$ does not divide.

• nice. Thanks for this. This is a good counter example.
– sku
Apr 13, 2017 at 1:11

It is not true. Only one of $n,n-1$ can be even, but one must be. That forces both $m-1,m+1$ to both be even. Say $n$ is even. It could have lots of factors of $2$ in such a way that $m-1$ and $m+1$ have to contribute factors to cover. Does this help find a counterexample?