Prove that $a^7-a$ is divisible by 168 when a is odd

Question

so I saw a similar question that proves $168\mid(a^6-1)$ when $(42,a) = 1$. But for this problem I was not given that gcd$(a,42)=1$. When I factor out a I get $168\mid a\cdot(a^6 - 1)$ and since $a$ is odd, $168 \nmid a$ so does that mean to prove that $168 \mid(a^6 - 1)$ when $a$ is odd? I looked into the numbers but I could find multiple examples where this is not the case. Thanks

Answer 1

As $a$ is odd, it is definitely not divisible by $168$.

But it does not necessarily imply that $(168,a)=1$ so that $168$ will have to divide $a^2-1$ as $168(=7\cdot8\cdot3)$ is not prime and $7$ and/or $3$ can divide odd $a$. See this .

We can follow the following method to establish the proposition:

using Fermat's Little Theorem $$a^p-a\equiv0\pmod p$$ for any prime $p$

As $\displaystyle168=7\cdot8\cdot3,$

$$p=7\implies a^7-a\equiv0\pmod7$$

$$p=3\implies a^3-a\equiv0\pmod3$$

$$\text{But }a^7-a=a(a^6-1)=a(a^2-1)(a^4+a^2+1)=\underbrace{(a^3-a)}(a^4+a^2+1)$$

For odd $a,$ $$a^2=(2b+1)^2=8\frac{b(b+1)}2+1\equiv1\pmod 8$$

$$\text{But }a^7-a=a(a^6-1)=a\underbrace{(a^2-1)}(a^4+a^2+1)$$

Answer 2

Theorem $\ $ if $\,p\ne q\,$ are odd primes then $\,p\!-\!1,q\!-\!1\mid n\!-\!1\,\Rightarrow\,8pq\mid a^n-a\ $ if $\,(a,2pq)=1$

Proof $\,\ {\rm mod}\ 8\!:\ a^n = a (a^2)^{{\frac{n-1}{2}}}\equiv a\ $ and ${\rm mod}\ p\!:\ a^n = a (a^{p-1})^{{\frac{n-1}{p-1}}}\equiv a\ $ and ditto mod $\,q.$

Answer 3

Alternatively, let $a$ be odd, and compute $(a^7-a)\mbox{ mod } m$ for $m=3$, $7$, and $8$.

First consider $m=3$. If $a\equiv_30$, $a^n\equiv_30$ for all $n>1$; for $a\equiv_31$, $a^n\equiv_31$ for all $n>1$. For $a\equiv_32$, $a^{odd}\equiv_32$ and $a^{even}\equiv_31$. In all these cases, $a^7\equiv_3 a$ and $3|(a^7-a)$..

For $m=7$: $a^7\mbox{mod} 7=(a\mbox{ mod } 7)^7$, and it can be checked that $(a^7\equiv_{\,7}a)$ for $0 \le a < 7$, whence $7|(a^7-a)$.

For $m=8$, note that $(odd)^2\equiv_81$, so $a^7\equiv_8a(a^2)^3\equiv_8 a$, and $7|(a^7-a)$.

Thus when $a$ is odd, $a^7-a$ is divisible by 3, 7, and 8, so it is divisible by $\mbox{lcm}(3,7,8)=168.$

Answer 4

You can also take a naive approach:

$$a^7-a=a(a^6-1)=a(a^3-1)(a^3+1)=(a-1)a(a+1)(a^2-a+1)(a^2+a+1)=$$

$$=(a-1)a(a+1)(a-3)(a-5)(a-2)(a-4)\pmod 7\;\;(**)$$

Now, $\;a\;$ is odd means exactly one of $\;a-1\;,\;a+1\;$ is divisible by $\;4\;$ and the other is divisible by two, so we already have divisibility by $\;4\cdot 2=8\;$

Also, exactly one of $\;a-1\,,\,a\,,\,a+1\;$ is divisible by $\;3\;$ , so we've already divisibility by $\;8\cdot3=24\;$

Finally, if $\;a=0,\pm 1\pmod 7\;$ then $\;(a-1)a(a+1)\;$ gives us divisibility by $\;7\;$ , otherwise $\;a=2,3,4\;\text{ or }\; 5\pmod 7\;$, and thus one of the last four factors above in (**), $\;a-2\;,\;a-3\;,\;a-4\;,\;a-5\;$ give us divisibility by $\;7\;$ , and get divisibility by $\;24\cdot7=168\;$.