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