<s>[INST] Is there any natural number between $(N^2 + N)$ and $(N^2 + 1)$ that can divide $(N^2 + N)\times(N^2 + 1)$ My question is fairly simple and explained in title. I'm trying to prove that there are no natural number(s) between $(n^2+n)$ and $(n^2+1)$ that can divide $(n^2+1) \times (n^2 + n)$
[EDIT]
I was trying to solve the problem $n^4 + n^3 + n^2+n+1 = a^2$.
I said that $n^4 + n^3 + n^2+n = (a-1)\times(a+1)$. Refactoring left side, i got:
$(n^2+n)\times(n^2+1) = (a-1)+(a+1)$. It is obvious that $a-1$ and $a+1$ are the closest dividers of $a^2-1$ so i need to prove that, in natural numbers, $n^2+n$ and $n^2+1$ are the closest dividers of $a^2-1$ so i can assume that $(n^2+n) = a+1$ and $(n^2+1) = a-1$ and continue to solve the problem in proper way. Thanks for any help or hint in front.
 [/INST] A proof by contradiction. Assume there is a $n^{2}+1+k|(n^{2}+n)(n^{2}+1)$
where $1\leq\,k\,\leq\,n-1$. Then
$n^{2}+k+1|(n^{2}+n)(n^{2}+k-k+1)$ hence $n^{2}+k+1|(n^{2}+n)k$ and then
$n^{2}+k+1|(n^{2}+k+1-k-1+n)k$ hence $n^{2}+k+1|-(k+1-n)k$ and the latter
is positive ($k=n-1$ is not acceptable since it gives $n^{2}+k+1=n^{2}+n$)
which implies that
$n^{2}+1+k\,\leq\,-k^{2}-k+nk$ and hence
$(n-\dfrac{k}{2})^{2}+3\dfrac{k^{2}}{4}+2k+1\,\leq\,0$
which is clearly a contradiction! So there is no such $k$ and the result is proved!
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