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

This is from a book. They have given the proof. But I didn't understand it well. I am looking for a simpler proof. Or it will be helpful if someone explain this a little bit more -
Proof: Let $m = n+1$ then we need to prove that $m(m-2) \mid 2^{(m-1)!}-1$. Because of $\phi(m) \mid (m-1)!$, we have $2^{\phi(m)} -1 \mid 2^{(m-1)!} -1$. And from Euler's theorem, $(m-2) \mid 2^{(m-1)!}-1$. Because $m$ is odd, $gcd(m,m-2)=1$ and the conclusion follows.

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    It seems to use properties from groups ? Do you know any ? – 2017-02-02
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    Which part don't you understand? I think this is more about what background you are lacking than the proof being not clear. It seems well-written to me! – 2017-02-02
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    @Maman No I dont ... :( I expected a elementary proof. – 2017-02-02
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    @PatrickDaSilva I just understood the first line ... But didn't understand any of the other lines ... from where they came ... A little bit more explanation will be sufficient – 2017-02-02
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    @RezwanArefin : This is very elementary! It only uses the Euler totient function, Euler's theorem and elementary properties of divisibility. You should learn about them if you don't know! (By the way, the verb is "to prove", but the noun is "a proof".) – 2017-02-02
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    Without stuff on groups it's actually more technical – 2017-02-02
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    @PatrickDaSilva I know all of them.. Euler theorem, Totient function... But how they applied here is not clear to me – 2017-02-02
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    If gcd(a,n)=1 then $a^{\phi(n)} \equiv 1 \pmod n$ – 2017-02-02
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    @Maman I know that :| – 2017-02-02
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    How $\phi(m) \mid (m-1)!$ ? – 2017-02-02
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    How is $\phi(m)$ when $m$ isn't prime ? – 2017-02-02
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    @Maman $\phi(m) = m(1 - \frac{1}{p_1})(1-\frac{1}{p_2}) ( \cdots )$? :| – 2017-02-02
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    Do you know in term of elements what does $\phi(m)$ represent ? – 2017-02-02
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    Didn't get you ... I didn't understand why $\phi(m) \mid (m-1)!$ :( – 2017-02-02
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    Because $\phi(m) < m$ or $\phi(m) \leq m-1$ and $(m-1)!=1\cdot 2\cdot 3\cdot ... \cdot (m-2) \cdot (m-1)$ – 2017-02-02

3 Answers 3

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Step 1: $\phi(m) |(m-1)!$

This is easy, since $\phi(m) \leq m-1$ therefore, it is one of the terms appearing in $(m-1)!$.

Step 2: If $a|b$ then $2^a-1|2^b-1$.

This follows from the fact that $b=ak$ and $$x^k-1=(x-1)(1+x+x^2+..+x^{k-1})$$ Replace $x$ by $2^a$.

Step 3: $2^{\phi(m)} -1 \mid 2^{(m-1)!} -1$.

Follows now from step 1 and step 2.

Step 4: $m|2^{\phi(m)}-1$

This is just the Euler Theorem.

Step 5: $m| 2^{(m-1)!} -1$

Comes from step 3 and step 4.

Step 6: $m-2|2^{(m-1)!} -1$.

Since Step 5 holds for all integers, it holds if we replace $m$ by $m-2$. Therefore $$m-2| 2^{(m-3)!} -1$$

By Step 3, $2^{(m-3)!} -1|2^{(m-1)!} -1$.

Step 7: If $a|c, b|c$ and $gcd(a,b)=1$ then $ab|c$.

Setting $a=m, b=m-2$ and $c=2^{(m-1)!}-1$ you get the claim.

This is a standard result in number theory, which can be viewed via prime factrisation or Extended Euclid algorithm.

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    Great ! I asked for this kind of explanation :) Thanks :D – 2017-02-02
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$2\,$ is coprime to $\,n\!+\!1\,$ so $\,{\rm mod}\ n\!+\!1\!:\, $ $\,2$ has order $\le n,\,$ i.e. $\,\color{#c00}{2^{\large k}\! \equiv 1}\,$ for $\,k\le n,\,$ thus $\,k\mid n!\:$ hence $\,2^{\large n!}\!\equiv (\color{#c00}{2^{\large k}})^{\large n!/k}\!\equiv 1.\,$ Similarly $\,2^{\large n!}\!\equiv 1\pmod{\!n\!-\!1}.$ Thus $\,2^{\large n!}\!-1\,$ is divisible by $\,n\!+\!1\,$ and $\,n\!-\!1\,$ hence also by their lcm = product, since $\:\gcd(n\!+\!1,n\!-\!1) = \gcd(n\!+\!1,2) = 1,\,$ by $n$ even.

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I have a shorter proof, because $n$ is even then $n^2-1$ is odd and $\gcd(n^2-1,2)=1$, thus according to Euler's theorem $$2^{\varphi(n^2-1)}\equiv 1 \pmod{n^2-1}$$ But totient function is multiplicative and $\gcd(n-1,n+1)=1$ or $$\varphi(n^2-1)=\varphi(n+1)\cdot \varphi(n-1)\leq n\cdot (n-2)