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Question - Prove that for any even positive integer $n, n^{2}-1$ divides $2^{n !}-1$

Proof: Let $m=n+1 .$ We need to prove that $m(m-2)$ divides $2^{(m-1) !}-1$

Because $\varphi(m)$ divides $(m-1) !$

we have $\left(2^{\varphi(m)}-1\right) |\left(2^{(m-1) !}-1\right)$

and from Euler's theorem, $m |\left(2^{\varphi(m)}-1\right) .$ It follows that $m |\left(2^{(m-1) !}-1\right) .$

Similarly $(m-2) |\left(2^{(m-1) !}-1\right) .$ Because $m$ is odd, gcd $(m, m-2)=1$ and the conclusion follows.

I just want to understand how they write

Similarly $(m-2) |\left(2^{(m-1) !}-1\right) .$

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Here's my initial thought:

Because $\varphi(m-2)$ divides $(m-3)!$ then it also divides $(m-1)!$.

Does that help?

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  • $\begingroup$ but we have to prove $(m-2) |\left(2^{(m-1) !}-1\right)$.. $\endgroup$ – Ishan May 15 at 8:49
  • $\begingroup$ ohh, yess i see that thanks.. $\endgroup$ – Ishan May 15 at 8:51
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Firstly we write the even integer $n$ as $m-1$, then $n^2-1=(n+1)(n-1)=m(m-2)$, where $m$ is odd. The totient funcion $\varphi(m)\leq m-1$ and is equal to $m-1$ if and only if $m$ is a prime. Hence $\varphi(m)\mid (m-1)!$, as $\varphi(m)$ has to be one of the numbers between $1$ and $m-1$. Hence $(m-1)!=k\varphi(m)$ for some $k\in\mathbb{Z}$. The factoring \begin{align*} 2^{(m-1)!}-1&= 2^{k\varphi(m)}-1\\ =(2^{\varphi(m)}-1)&(2^{\varphi(m)(k-1)}+ 2^{\varphi(m)(k-2)}+\dotsb+ 2^{\varphi(m)}+ 1) \end{align*} shows we have $(2^{\varphi(m)}-1)\mid (2^{(m-1)!}-1)$, and by Euler's theorem $2^{\varphi(m)}\equiv1\pmod{m}$, implying $m\mid (2^{(m-1)!}-1)$ also.

Similarly $\varphi(m-2)\mid (m-1)!$, as $\varphi(m-2)$ has to be one of the numbers between $1$ and $m-2$. Hence $(m-1)!=j\varphi(m-2)$ for some $j\in\mathbb{Z}$. The factoring \begin{align*} 2^{(m-1)!}-1&= 2^{j\varphi(m-2)}-1\\ =(2^{\varphi(m-2)}-1)&(2^{\varphi(m-2)(j-1)}+ 2^{\varphi(m-2)(j-2)}+\dotsb+ 2^{\varphi(m-2)}+ 1) \end{align*} shows we have $(2^{\varphi(m-2)}-1)\mid (2^{(m-1)!}-1)$, and by Euler's theorem $2^{\varphi(m-2)}\equiv1\pmod{m-2}$, implying $m-2\mid (2^{(m-1)!}-1)$ also.

Finally since $m$ and $m-2$ are adjacent odd numbers, they are also coprime, which shows the product $m(m-2)=(n+1)(n-1)=n^2-1\mid 2^{n!}-1$ as required.

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