Prove that $14322\mid n^{31} - n$

Question

Im trying to prove that $ 14322 \mid n^{31} - n $, $ \forall n \in \mathbb{Z}$

My thought was to rewrite $n^{31} - n$ which a got to $$ 2(n-1)\sum_{i=0}^{n} i \sum_{k=0}^{14} n^k \sum_{j=0}^{14} (-1)^kn^k $$ but this didnt (in my opinion) contribute to much.

I tried to assume that it was false, and then try to get a contradiction.

saying that $n^{31}-n=q\cdot 14322 + r$, $0<r<q, q\in \mathbb{Z}$ then reaching a contradiction $\Rightarrow$ $r=0$. I did not see any obvious contradiction.

Hints?

Elaqqad · Accepted Answer · 2015-04-09T18:57:40.527

10

The usual method to solve this is to use Fermat's Little Theorem: $$14322=2\cdot3\cdot7\cdot11\cdot 31 $$

and you prove that $n^{31}-n$ is divisible by each prime in this factorization

Edit As noticed in the comment we have $n^{p}-1$ divides $n^{pq}-1$

Fermat implies that $11$ divides $n^{11}-n$, and as $n^{10}-1$ divides $n^{30}-1$ then $n^{11}-n$ divides $n^{31}-n$ hence $11$ divides $n^{31}-n$
As $n^{6}-1$ divides $n^{30}-1$ and then $n^{7}-n$ divides $n^{31}-n$ gives you the divisibility by $7$ and you can do the same for others

edited Apr 09 '15 at 18:57

answered Apr 09 '15 at 18:02

Elaqqad

13,725

you meant $n^{31}-n$ – Dr. Sonnhard Graubner Apr 09 '15 at 18:04
@Dr.SonnhardGraubner, yes, that's exactly what he wrote! – Prasun Biswas Apr 09 '15 at 18:05
Oh. It become alot more easier when I googled fermats Little Theorem... – Olba12 Apr 09 '15 at 18:08
now yes he has corrected it – Dr. Sonnhard Graubner Apr 09 '15 at 18:08
it is all ok with your syntax – Dr. Sonnhard Graubner Apr 09 '15 at 18:14
Hint: use $n^{pq}-1=(n^p-1)(n^{pq-p}+\cdots+1)$ and so $n^p-1\mid n^{pq}-1$ for any natural $p,q$. – user89167 Apr 09 '15 at 18:27
I dont see how I can use that – Olba12 Apr 09 '15 at 18:40
1

Do you mean since $n^{31}-n = n(n^{15}-1)(n^{15}+1)$? – Olba12 Apr 09 '15 at 18:43
Yeah so by fermat its divisable by 31, but how do I prove the others? – Olba12 Apr 09 '15 at 18:50
@Olba12 see the update – Elaqqad Apr 09 '15 at 18:58
I don't think you need to duplicate efforts with e.g., $n^6-1\to n^7-n$, since these are primes we have $n^6\equiv 1\pmod 7$... It seems to confuse things a bit since you are re-using the single $n$ factor in multiple places without telling us that we don't need $n^5$ to make it all work. – abiessu Apr 09 '15 at 19:01
@abeiesu I used it like that to avoid using $\gcd(n,7)=1$ in multiple places – Elaqqad Apr 09 '15 at 19:10
I liket the thing you added, thanks. It made everything clear. :) – Olba12 Apr 09 '15 at 19:30

Bill Dubuque · Answer 2 · 2015-04-09T19:34:39.423

If you don't know modular arithmetic then you can use little Fermat and the following

$\ \ p\mid n^p\!-\!n\mid n^{31}\!-\!n\,$ by $\,n^{\large \color{#c00}{p-1}}\!-\!1\mid n^{\color{#c00}{30}}\!-\!1\,$ by $\,\color{#c00}{p\!-\!1\mid 30}\,$ for all $\, p\mid 14322=2\cdot3\cdot7\cdot11\cdot 31$

Otherwise, the easy direction $\,\color{#90f}{(\Leftarrow)}\,$ of the following theorem applies.

Theorem $ $ (Korselt's Carmichael Criterion) $\ $ For $\rm\:1 < e,n\in \Bbb N\:$ we have

$$\rm \forall\, a\in\Bbb Z\!:\ n\mid a^e\!-a\ \iff\ n\ \ is\ \ squarefree,\ \ and \ \ \color{#c00}{p\!-\!1\mid e\!-\!1}\ \, for\ all \ primes\ \ p\mid n\quad $$

Proof $\ \ \color{#90f}{(\Leftarrow)}\ \ $ Since a squarefree natural divides another natural iff all its prime factors do, we need only show $\rm\: p\mid a^e\!-\!a\:$ for each prime $\rm\:p\mid n,\:$ or that $\rm\:a \not\equiv 0\:\Rightarrow\: a^{e-1}\equiv 1\ \ ( mod\ p),\:$ which, since $\rm\:p\!-\!1\mid e\!-1,\:$ follows from $\rm\:a \not\equiv 0\:\Rightarrow\: \color{#c00}{a^{p-1} \equiv 1}\ \ ( mod\ p)\:$ by little Fermat, i.e.

$$\rm \color{#c00}{e\!-\!1 = (p\!-\!1)}k\ \Rightarrow\ a^{\large e-1} \equiv (\color{#c00}{a^{\large p-1}})^{\large k}\equiv \color{#c00}1^{\large k}\equiv 1\!\!\!\pmod p \qquad\qquad $$

$(\Rightarrow)\ \ $ Given that $\rm\: n\mid a^e\!-\!a\:$ for all $\rm\:a\in\Bbb Z,\:$ we must show

$$\rm (1)\ \ n\,\ is\ squarefree,\quad and\quad (2)\ \ p\mid n\:\Rightarrow\: p\!-\!1\mid e\!-\!1$$

$(1)\ \ $ If $\rm\,n\,$ isn't squarefree then $\rm\,1\neq \color{#0a0}{a^2}\!\mid n\mid \color{#0a0}{a^e}\!-\!a \Rightarrow\: a^2\mid a\:\Rightarrow\!\Leftarrow$ $\rm\: (note\ \ e>1\: \Rightarrow\: \color{#0a0}{a^2\mid a^e})$

$(2)\ \ $ Let $\rm\ a\ $ be a generator of the multiplicative group of $\rm\:\Bbb Z/p.\:$ Thus $\rm\ a\ $ has order $\rm\:p\!-\!1.\:$ Now $\rm\:p\mid n\mid a\,(a^{e-1}\!-\!1)\:$ but $\rm\:p\nmid a,\:$ thus $\rm\: a^{e-1}\!\equiv 1\,\ ( mod\ p),\:$ therefore $\rm\:e\!-\!1\:$ must be divisible by $\rm\:p\!-\!1,\:$ the order of $\rm\,\ a\,\ (mod\ p)$.

abiessu · Answer 3 · 2015-04-09T19:46:30.313

2

One way to solve this is to break down the polynomial $n^{31}-n$ directly, taking into consideration that $x-1\mid x^q-1$:

$$\begin{align}n^{31}-n&=n(n^{30}-1)\\ &=n(n^{10}-1)(n^{20}+n^{10}+1)\\ &=n(n^6-1)(n^{24}+n^{18}+n^{12}+n^6+1)\\ &=n(n^2-1)(n^{28}+n^{26}+\dots+1)\\ &=n(n-1)(n^{29}+n^{28}+\dots+1) \end{align}$$

Since $14322=2\cdot3\cdot7\cdot11\cdot 31$, we can use Fermat's Little Theorem on all the values $2,3,7,11,31$ with the terms $n^2-n,n^3-n,n^7-n,n^{11}-n, n^{31}-n$. Note that we do not need a factor $n^5$ to account for all of these divisibility conditions, since $p\mid n$ is the only time when the $n$ factor is needed in $n(n^{p-1}-1)$, meaning that when $14322\mid n$, we get $14322\mid n^{31}-n$ from the single $n$ factor, and when $(14322,n)\lt 14322$ all the factors of $14322$ which are not present in $n$ appear in $n^{30}-1$.

edited Apr 09 '15 at 19:46

answered Apr 09 '15 at 18:59

abiessu

8,115

1

$p\mid n^p!-!n\mid n^{31}!-!n,$ by $,n^{\large \color{#c00}{p-1}}!-!1\mid n^{\color{#c00}{30}}!-!1,$ by $,\color{#c00}{p!-!1\mid 30},$ for all $, p\mid 14322=2\cdot3\cdot7\cdot11\cdot 31\ $ $\tag*{}$ This is a divisibility form of the congruence argument in my answer. – Bill Dubuque Apr 09 '15 at 19:26
Very true... I hope that the vertical alignment makes it a little clearer, but perhaps I could do something else to improve my answer... – abiessu Apr 09 '15 at 19:38
1

Yes, the alignment makes it clearer. Note that I added the expression in the comment only to help readers understand the relationship between the various methods. But if that motivates you to polish your answer then all the better! The site would be much richer if there were more collaboration. – Bill Dubuque Apr 09 '15 at 19:43
1

Note $,n^J!-!1\mid n^{JK}!-!1, $ is $\ a!-!1\mid a^K!-!1, $ for $,a = n^J,,$ so we don't need to compute the quotient (cofactor) to prove that the divisibility is true. That's one of the major advantages of modular methods, e.g. $, {\rm mod}\ n^J!-!1!:\ \color{#c00}{n^J\equiv 1},\Rightarrow, n^{JK}\equiv (\color{#c00}{n^J})^K\equiv \color{#c00}1^K\equiv 1\ \ $ – Bill Dubuque Apr 09 '15 at 19:58

Prove that $14322\mid n^{31} - n$

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