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Show that for any $a \in \mathbb{Z}, 42 \mid (a^{7} − a)$.

I saw this question on Rosen textbook and it doesn't have answer key so I am wondering can you guide me how to do it?

What I have tried is that since I don't know the number a so I substitute a number for a.

Bill Dubuque
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guest11
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4 Answers4

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$$a^7-a=a(a-1)(a+1)(a^2-a+1)(a^2+a+1).$$ Since it has a product of three consecutive integers as a factor, it is divisible by $3!$.

By Fermat's little theorem, it is divisible by $7$.

Since $\gcd(6, 7)=1,$ this expression is divisible by $6\times7=42.$

Bumblebee
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  • why can't I just have 3 of $(a^2+a+1)$ – guest11 Nov 11 '15 at 04:15
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    @guest11 What Nilan means, which may still be a little unclear, is that if $a$ was any number, then one of those three factors is divisible by $3$. Consider three cases: $a = 3n$ for some $n$, $a = 3n + 1$ for some $n$, and $a = 3n + 2$ for some $n$. In the first case, the factor $a$ is divisible by three. In the second case, $(a - 1)$ is divisible by three. In the third case, $(a + 1)$ is divisible by three. No matter what, one of those factors will be divisible by three. Similar logic shows you that one of those factors is divisible by two. – Axoren Nov 11 '15 at 04:26
  • @Axoren There is a nice easy proof of any $n$ consecutive number is divisible by $n!$: Consider the number ${a\choose n}$ where $a$ is the largest number in the consecutive numbers. This is by definition the number of ways to choose and must be an integer. – cr001 Nov 11 '15 at 04:31
  • @cr001 That is a sneaky proof and I really like it. Though easy in hindsight, it took me a handful of minutes to grasp and accept it. In this case, we could have had to use $\binom{a + 1}{n}$ as it would have been the largest. – Axoren Nov 11 '15 at 04:37
  • Yes, in this case it would be "Since ${a+1\choose3}$ is an integer $3!$ divide $(a-1)a(a+1)$". I was talking about the general case and sorry for using the same variable $a$ as in the question. – cr001 Nov 11 '15 at 04:39
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(1) If $a$ is multiple of $7$ then $a^7-a$ is a multiple of $7$; if $\gcd(a, 7)=1$ then by FLT $a^6\equiv1\pmod{7}$ hence $a^7-a=a(a^6-1)$ is a multiple of 7.

(2) If $a$ is multiple of $3$ then $a^7-a$ is a multiple of $3$; if $\gcd(a, 3)=1$ then by FLT $a^2\equiv1\pmod{3}$ hence $a^7-a=a(a^2-1)(a^4+a^2+1)$ is a multiple of 3.

(3) If $a$ is multiple of $2$ then $a^7-a$ is a multiple of $2$; if $\gcd(a, 2)=1$ then by FLT $a\equiv1\pmod{2}$ hence $a^7-a=a(a-1)(a+1)(a^4+a^2+1)$ is a multiple of 2.

Axoren
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cr001
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Hint : $a = b \mod p$ implies that $ a^n-a = b^n-b \mod p$ Use it with $p=2,3,7$ (as $42=2*3*7$) and all the possible $b$ (ie $0,1,..,p-1$)

Axoren
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stity
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$42 = 2\cdot3\cdot7$, so we can check individually if

$2|a^7-a$ and $3|a^7-a$ and $7|a^7-a$

simultaneously hold. For the first one, we consider two cases: either $a$ is odd or is even. This means that either $a\equiv0\pmod 2$ or $a\equiv1\pmod2$.

Thus, we either have $1^7-1\equiv 0\pmod2$ or $0^7-0\equiv0\pmod2$. In both case, we can conclude that $2|a^7-a$.

Next we deal with 3 cases: either $a\equiv0\pmod3$ or $a\equiv1\pmod3$ or $a\equiv2\pmod3$. Plug the $a$ in any of the three cases and you should get that $a^7-a\equiv0\pmod3$

Finally you deal with 7 cases the same way around.

EA304GT
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