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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.

Bill Dubuque
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seymayneym
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    Not for small $n$. What have you tried? – Henry Sep 20 '22 at 15:33
  • Hint: Otherwise there exists $a,b$ such that $n^2+1<a\le b<n^2+n$ and $ab=(n^2+1)(n^2+n)$. We must also have $a+b<(n^2+1)+(n^2+n)$. Do you see why?. What can you say about $(a+b)^2-4ab$? – Jyrki Lahtonen Sep 20 '22 at 16:04
  • I have tried, with python, until $10^5$. And found nothing! But i want to prove it formally. – seymayneym Sep 20 '22 at 16:04
  • I can expand that, but please check out our guide for new askers, and try to improve the question with pieces of context, the origin of the question, or partial work (how many values of $n$ have you checked, or some combination thereof. – Jyrki Lahtonen Sep 20 '22 at 16:07
  • @JyrkiLahtonen Why $ab=(n^2+n)(n^2+1)$? – nonstudent Sep 20 '22 at 16:11
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    @nonstudent Let $a$ be that factor, and define $b=(n^2+1)(n^2+n)/a$. We can swap the roles of $a$ and $b$ according to which is larger. – Jyrki Lahtonen Sep 20 '22 at 16:20
  • See here, and links in there, for more discussion on the problem you started with. The principle too close to a square to be square in Jack D'Aurizio's answer is worth remembering (and actually how I was thinking about your problem as well). – Jyrki Lahtonen Sep 20 '22 at 17:23

2 Answers2

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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.

Jyrki Lahtonen
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    Undoubtedly there is a well known general bound on how "close" can two factorizations of a given number $N$ be. If $AB=N=UV$ then $$(A+B)^2+(A-B)^2=4N=(U+V)^2+(U-V)^2.$$ So if $|A-B|<|U-V|$ we must also have $|A+B|<|U+V$ et cetera. I could not find a useful dupe, but may be I used wrong buzzwords. – Jyrki Lahtonen Sep 20 '22 at 16:55
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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!