How can I prove that

If $n$ is a positive integer such that $$n>8$$ then $$\frac{n!-1}{2n+7}$$ is never an integer? Some of the first things that came to my mind is that $n!-1$ is not divisible by all numbers from $2$ to $n$, so if $$2n+7<n^2$$ That is, $n>3$

then $2n+7$ must be a prime number in order to actually get an integer.

But thats all i have got until now, after checking some basic number theory theorems, I even tried linking it to Wilson´s theorem, unsuccesfully. Any hint or idea will be very appreciated


2 Answers 2


As you observed, unless $2n+7$ is prime, you get $2n+7|n!$. This is clear, since if $2n+7$ is not prime, you can write it as $2n+7=ab$ and since neither $a$ nor $b$ are 2, both must be less than $n$. As Jonas pointed, if $a \neq b$ the conclusion is immediate, while if $a=b$ we have $a^2=2n+7$ and one can easily conclude that $2a \leq n$.

So lets cover the case $p=2n+7$ is prime. Then $n=\frac{p-7}{2}=\frac{p-1}{2}-3$.

Then you get

$$n! \equiv 1 \pmod p$$


$$(\frac{p-1}{2}-3)! \equiv 1 \pmod p \,.$$ $$(\frac{p-1}{2})! \equiv \frac{p-5}{2}\frac{p-3}{2}\frac{p-1}{2} \equiv 8^{-1}(-15) \pmod p \,.$$

Square both sides:

$$[(\frac{p-1}{2})!]^2 \equiv 64^{-1}(15)^2 \pmod p \,.$$

The LHS is exactly $(p-1)! (-1)^{\frac{p-1}{2}}$. Thus

$$(-1)^{\frac{p+1}{2}} \equiv 64^{-1}(15)^2 \pmod p \,,$$


$$\pm 64 \equiv 225 \pmod p$$

This means that $p$ is a divisor of $225 \pm 64$. Note that $p=2n+7>23$.

This reduces the problem to few cases to check.

  • $\begingroup$ @Jonah good point. The case $a=b$ is trivial though. $\endgroup$
    – N. S.
    Commented Jun 7, 2013 at 4:02

Hint: $\ $ Supposing that $\ p = 2n\!+\!7\ $ is prime and $\,\color{#0a0}{p\mid n! - 1},\,$ and applying Wilson's Theorem

$\begin{eqnarray} {\rm mod}\ \ p\!:\ {-}1 \equiv (p\!-\!1)! &\equiv&\, (2n\!+\!6)\!&&\!(2n\!+\!5)\!\!&&\cdots (n\!+\!7) &&(n\!+\!6)\cdots (n\!+\!1)\, n!\\ &\equiv& \ \ \ (-1)&&\ \ (-2)&&\cdots\ \, (-n)&&(n\!+\!6)\cdots (n\!+\!1)\, n! \\ &\equiv& && && \quad\ \ \ (-1)^n &&(n\!+\!6)\cdots (n\!+\!1)\ \ \ \ {\rm by}\ \ \ \color{#0a0}{n!\equiv 1} \end{eqnarray}$ $\begin{eqnarray}\ \stackrel{\large \times\ 2^6}\Rightarrow\ \pm 2^6 &\equiv&\, (\color{#c00}{2n}\!+\!12)(\color{#c00}{2n}\!+\!10)\cdots (\color{#c00}{2n}\!+\!2)\\ &\equiv&\, (5)\,(3)\,(1)\,(-1)\,(-3)\,(-5)\equiv -15^2\ \ \ {\rm by}\ \ \ \color{#c00}{2n}\equiv -7\ \ \, ({\rm mod}\ \ p = 2n\!+\!7)\\ \\ \Rightarrow\ \ \ \pm64 &\equiv&\, 225,\ \ \ {\rm i.e.}\ \ \ p\mid 225\pm 64\\ \end{eqnarray}$


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