Prove: if $n\mid 7^n+6^n$ and $n>1$, then $13\mid n$

Question

Prove: if $n\mid 7^n+6^n$ and $n>1$, then $13\mid n$

Let $p$ be the least prime number such that $p\mid n$.

And I want to show that $p=13$

Let $d$ be the least number such that: $14^d\equiv 0 \pmod {p}$

And by Fermat's little theorem I have: $14^{p-1}\equiv 0 \pmod{p}$

Here I'm stuck

Answer 1

$n\mid 7^n\!+6^n\!=:a_n$ odd $\Rightarrow \color{#0af}{n\ \rm odd}$, so $\,a_n = 7^n\!\!\color{#0af}{-\!(-6)^n},\,$ so below $\,b,c= 7,-6\,\Rightarrow\, \color{#c00}{p = 13}$.

Lemma $\, $ If $\,(b,c)\!=\!1\ $ and $\ 1<n\mid b^n\!-c^n$ then $\,p\mid b\!-\!c\,$ for $\,\color{#c00}{p = {\rm least}}$ prime factor of $\,n$.

Proof $\,\ \color{#f60}{p\nmid b}\,$ else $\,p\mid n\mid b^n\!-c^n\Rightarrow\,p\mid c,\,$ contra $\,(b,c)\!=\!1,\,$ so $\,\color{#90f}{p\nmid c}\,$ by symmetry. If $\ \color{#0a0}{p\nmid b\!-\!c}\ $ $\rm\color{#90f}{then}$ $\!\bmod p\!:\ a:= \frac{b}c\!\not\equiv \color{#0a0}1,\color{#f60}0,$ and $\, a^n\equiv 1,\,$ thus $\, {\rm ord}(a)\mid n\Rightarrow {\rm ord}(a)\ge\color{#c00}p,\,$ contra $\ a^{p-1}\equiv 1\,$ by little Fermat $\,({\rm ord}(a)\neq 1$ by $\,\color{#0a0}{a\neq 1}\,$ so the least value $\,{\rm ord}(a)\,$ can take is least factor $> 1\,$ of $\,n,\,$ which is its least prime factor $\,\color{#c00}p).$

Answer 2

Let $\,p\,$ be the least prime divisor of $\,n$

Observe $\,p\mid(6^n+7^n)(7^n-6^n)\,$ and little Fermat to obtain:

$p\mid 7^{2n}-6^{2n},\, 7^{p-1}-6^{p-1}\iff\, p\mid (7^{2n}-6^{2n},\, 7^{p-1}-6^{p-1})$

By below theorem and $\,(2n,p-1)=2\, $ we get $\,p\mid 7^2-6^2=13$

Theorem: $\,(a,b)=1,\,a>b\,\,\Rightarrow\,\, (a^m-b^m,a^n-b^n)=a^{(m,n)}-b^{(m,n)}$

Proof: Use $\,x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+\cdots+xy^{k-2}+x^{k-1})\,$

and $\gcd$ definition $n\mid a,b\iff n\mid (a,b)$ to prove:

$a^{(m,n)}-b^{(m,n)}\mid a^m-b^m,\, a^n-b^n\iff$

$a^{(m,n)}-b^{(m,n)}\mid (a^m-b^m,a^n-b^n)=: d\ \ \ (1)$

$a^m\equiv b^m,\, a^n\equiv b^n$ mod $d$ by definition of $d$.

Bezout's lemma gives $\,mx+ny=(m,n)\,$ for some $x,y\in\Bbb Z$.

$(a,b)=1\iff (a,d)=(b,d)=1$, so $(a^m)^x,(b^n)^y$ mod $d$ exist.

$a^{(m,n)}\equiv (a^{m})^x(a^n)^y\equiv (b^{m})^x(b^n)^y\equiv b^{(m,n)}\pmod{\! d}\ \ \ (2)$

$(1)(2)\,\Rightarrow\, a^{(m,n)}-b^{(m,n)}=d$

Answer 3

Here is a stronger claim.

Claim: $n\mid 7^n+6^n$ iff $n=13^k$.

The forward direction has already been answered. Now assume $n=13^k$ we show that $n\mid 7^n+6^n$. We use $p$adic valuation. Since $n$ is power of $13$ it is sufficient to prove that $$\nu_{13}(7^n+6^n)\ge \nu_{13}(n)$$

LTE gives $\nu_{13}(7^n+6^n)=\nu(7+6)+\nu_{13}(13^k)=1+k\ge k=\nu_{13}(n)$

Answer 4

OP poses two questions: If $n\mid 7^n+6^n$, first show $13\mid n$, and second show that $13$ is the smallest prime factor of $n$

Start by noting that any prime factor of $n$ cannot be a prime factor of either $7^n$ or $6^n$, as a prime factor which divides $n$ and either of $7^n$ or $6^n$ must also divide the other. That is impossible because $\gcd(7^n,6^n)=1$. Thus, none of $2,3,7$ can be factors of $n$. In particular, this means that $n$ must be odd, a fact which is also apparent because the sum $7^n+6^n$ is odd.

Next, recall that for any odd exponent $k$, it is the case that $(a+b)\mid (a^k+b^k)$. Since $n$ is restricted to being odd, this means that $(7+6)\mid (7^n+6^n)$. This shows that $13\mid n$, answering OP's first query.

Since we know that $2,3,7$ are not divisors of $n$, we need only show that $5$ and $11$ are not divisors of $n$ to establish that $13$ is the smallest prime divisor of $n$

The case of $5$ is simple: $7^n+6^n \equiv 2^n+1^n \bmod 5$, and $2^n+1^n \equiv 0 \bmod 5 \iff n=4k+2$. That requires $n$ to be even, and $n$ is restricted to being odd, so $5\mid n$ is ruled out.

The case of $11$ takes a bit more work. Working modulo $11$, and recalling that when a modulus $p$ is prime, $a^k \bmod p \equiv a^{k \bmod (p-1)} \bmod p$ $$\begin {array}{c|c|c|c|c|c|c|c|c|c|c|c|} \\ n\equiv &1&2&3&4&5&6&7&8&9&10& \bmod 10 \\ 6^n \equiv &6&3&7&9&10&5&8&4&2&1& \bmod 11 \\ 7^n \equiv &7&5&2&3&10&4&6&9&8&1& \bmod 11 \\ \sum \equiv &2&8&9&1&9&9&3&2&10&2& \bmod 11 \\ \end {array}$$ Since the sum $7^n+6^n \not \equiv 0 \bmod 11$ for all $n$, it is the case that $11 \not\mid n$.

This shows that no prime smaller than $13$ divides $n$, answering OP's second query.