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.

  • $\begingroup$ 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? $\endgroup$ – Joppy Apr 12 '17 at 22:44
  • $\begingroup$ yes... tried examples. I feel it is true for non-prime m,n. But looking for proof. $\endgroup$ – sku Apr 12 '17 at 22:48
  • 1
    $\begingroup$ 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. $\endgroup$ – Badam Baplan Apr 12 '17 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.

  • $\begingroup$ nice. Thanks for this. This is a good counter example. $\endgroup$ – sku Apr 13 '17 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?


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