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Question:

show that: there exsit infinite postive integer $n$ such

$$n^3+1|n!$$

Buy the way,I know prove this this simple problem:

prove there are many infinte postive integer numbers $n$,such $n^2+1|n!$

This problem we only consider this pell equation $$n^2+1=5y^2$$ have a solution $(2,1)$, so the pell equation have many infite solution.

because $$2y=2\sqrt{\dfrac{n^2+1}{5}}<n\Longrightarrow n>5, 5,y,2y\in\{1,2,3,\cdots,n\}$$ so there exsit infinte $n$ such $$n^2+1|n!$$

But this is $n^3+1$ so I can't .Thank you

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    $\begingroup$ why $(n^2-n+1)|n!?$ $\endgroup$ – china math Aug 26 '14 at 9:05
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    $\begingroup$ Is there any reason that you believe this holds for infinitely many $n$ ? For such $n$ we need that $n^3+1$ has only "small" prime factors, e.g., as it is the case for $n=17$ or $n=31$. $\endgroup$ – Dietrich Burde Aug 26 '14 at 9:45
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I will prove there are infinitely many solutions of the form $n=3a^2$. For such $n$ we have $3 \nmid n+1$, hence $\gcd(n+1,n^2-n+1)=1$ and it is sufficient to show that both $n+1 \mid n!$ and $n^2-n+1 \mid n!$. We have $n+1 \mid n!$ whenever $n+1 = 3a^2+1$ is not prime, whereas $n^2-n+1 = 9a^4-3a^2+1$ factors as $(3a^2+3a+1)(3a^2-3a+1)$. Since both factors are relatively prime, we should have $3a^2+3a+1 \mid (3a^2)!$ and $3a^2-3a+1 \mid (3a^2)!$. The latter is always true, whereas $3a^2+3a+1 \mid (3a^2)!$ is true when $3a^2+3a+1$ is not prime. Therefore, it is sufficient to show that there are infinitely many $a$ for which neither $3a^2+1$ nor $3a^2+3a+1$ is prime, which should not be that hard (e.g. take $a \equiv 1 \mod 7$ and $a \equiv 2 \mod 13$).

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