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I'd like to solve this one similarly to my previous question: Is this a Valid proof for $(2n+1,3n+1)=1$?

I did find a somewhat related post that uses a different method: How to show that $\gcd(n! + 1, (n + 1)! + 1) \mid n$?

So how would I go about this? I can write the above like:

$\exists \ d \ \in \mathbb{Z}$

  1. $n!+1 \equiv 0$ (mod $d$)
  2. $(n+1)!+1 \equiv 0$ (mod $d$)

Not sure what to do next. Any hints?

Thanks!

Christoph
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5 Answers5

9

By the Euclidean Algorithm $$\left(n!+1,\ (n+1)!+1\right)=\left(n!+1,\ (n+1)!+1-(n!+1)\right)$$ $$=\left(n!+1,n\cdot n!\right).$$ It is clear the last two must be relatively prime, since if $p|\ (n!\cdot n)$ then $p| n!$ (and there is an extra +1 hanging around).

Eric Naslund
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7

Here's a purely equational proof. Simply put $\rm\ k = (n-1)!\ $ in

Theorem $\rm\ \ ((n+1)\, n\, k+1,\ n\, k+1)\ =\ 1$

Proof $\ \ $ Working modulo the gcd $\rm\: := d\:$ we have

$(1)\rm\quad\quad (n+1)\, \color{#c00}{n\, k}\ \equiv\: -1\quad\quad$ by $\rm\ d\ |\ (n+1)\, n\, k+1$

$(2)\rm\quad\quad\phantom{(n+1)\, } \color{#c00}{n\, k}\ \equiv\: -1\quad\quad$ by $\rm\ d\ |\ n\, k+1$

$(3)\rm\quad\quad\phantom{(n+1)\ n\ } n\ \equiv\,\ \ \ 0\quad\quad $ by substituting $\:\color{#c00}{(2)}\:$ in $\:(1)\:$

$(4)\rm\quad\quad\phantom{(n+1)\ n\ } 0\ \equiv\: -1\quad\quad$ by substituting $\:(3)\:$ in $(2)$

Thus we conclude $\rm\: 0\ \equiv\ 1,\, $ i.e. $\rm\ d\ |\ 1\quad$ QED

Unwinding the linear relations used in the above proof (or, equivalently, using the extended Euclidean algorithm) yields the Bezout relation that I gave in the question that you linked to, viz.

$$ \rm 1\ =\ (n-1)!\ ((n+1)!+1)\ +\ (1-(n+1)!/n)\ (n!+1)$$

Notice how what seems like magic viewed in terms of divisibility relations is reduced to a purely mechanical elimination process in equational form. In higher number theory you'll learn more precisely how linear algebra methods such as Gaussian elimination extend from fields to certain rings, e.g. google Hermite and Smith normal forms.

Bill Dubuque
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  • Hm, I don't see how n! is also congruent to -1. Could you expand on that? Thanks for the encouragement! I am just an engineer, so this one class will probably all the number theory I will learn, unless I end up inventing error correcting codes at my job. I am enjoying it though! – Christoph Mar 08 '11 at 03:44
  • Ah wait. $n! + 1 \equiv 0$ means $n! \equiv -1$? I can just do that, just as if the $\equiv$ were an =? That n! is congruent to both 0 and -1 makes no sense in my head though.... – Christoph Mar 08 '11 at 03:46
  • @Chris: Yes, you can add and multiply congruence just like integer equations. For an introduction see the Wikipedia article on modular arithmetic. – Bill Dubuque Mar 08 '11 at 03:53
4

Suppose $m$ is a positive integer dividing both $n! + 1$ and $(n+1)! + 1$. The goal is to show that $m = 1$. Note that $m$ also divides $(n+1)*(n! + 1) - ((n+1)! + 1)$, which is equal to $(n+1)! + (n+1) - (n+1)! - 1 = n$. Since $m$ divides $n$, it also divides $n!$. Since it divides $n! + 1$ as well, it divides $n! + 1 - n!$ or just $1$. Hence $m = 1$ as needed.

Zarrax
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  • Could you explain how you know that m divides (n+1)∗(n!+1)−(n+1)!+1? – Christoph Mar 08 '11 at 03:41
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    if $m$ divides $a$ and $b$ then $m$ divides $ka + lb$ for any integers $k$ and $l$. So here $a = n! + 1$ and $b = (n+1)! + 1$, while $k = (n+1)$ and $l = -1$. – Zarrax Mar 08 '11 at 03:52
  • You can also argue that two consecutive integers are coprime, so it suffices to prove m divides $n!$ for the prove to be done (since gdc is unique) – Eduardo V. Kuri Oct 07 '22 at 00:44
1

This is a direct application to the division algorithm which says:

for any two integers a,b, a>b the following holds: $$a = qb + r$$

If we show that $0<r<b$, we are done.

Replace $a$ by $(n+1)!+1$, $q$ by $n$, $b$ by $(n!+1)$ and $r$ by $n!-n+1$ which is greater than $0$ and less than $b$, as required.

More Clarificatoin:

The left hand side is:$$(n+1)!+1$$ while the right hand side is: $$n(n!+1)+(n!-n+1) \Longrightarrow nn!+n+(n!-n+1) \Longrightarrow n!(n+1)+1 \Longrightarrow (n+1)!+1$$

Maged Saeed
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1

This is just an alternative to the approach in Eric Naslund's answer.

From the rule $(a,b)=(a,b-ka)$ for any $k$ along with $(a,b)=(b,a)$ and $(a,b)=(|a|,|b|)$, we have

$$\begin{align} (n!+1,(n+1)!+1) &=(n!+1,(n+1)!+1-(n+1)(n!+1))\\ &=(n!+1,-n)\\ &=(n,n!+1)\\ &=(n,n!+1-(n-1)!n)\\ &=(n,1)\\ &=1 \end{align}$$

Barry Cipra
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