<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] Let $N=(n^2+1)(n^2+n)$. Assume that there exists an integer $A$ such that $A\mid N$ and $(n^2+1)<A<(n^2+n)$. Then the complementary factor $B=N/A$ is also in that interval.
Because $A$ and $B$ are closer to each other than $n^2+1$ and $n^2+n$, and yet $AB=(n^2+1)(n^2+n)$, we must have
$A+B<(n^2+1)+(n^2+n),$ or $A+B\le 2n^2+n$.
But then
$$(A-B)^2=(A+B)^2-4AB\le (2n^2+n)^2-4(n^2+1)(n^2+n)=-4n-3n^2,$$
where the right hand side is negative. This is a contradiction.
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