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Suppose we are given a number $a \in \mathbb{Z}$ prove that $42|a^7-a$.

I'm not too sure how to start any ideas?

Jack Frost
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James K
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5 Answers5

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Reduce $a^7-a$ mod 2,3, and 7 (the factors of 42) and show in each case that you get 0, so that 2,3, and 7 are all factors of $a^7-a$

Ryan Vitale
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Hint $\ $ It follows immediately from a slight generalization of $\color{#c00}{\rm Fermat's}$ little Theorem:

  • $ $ If $\ p\!-\!1\mid e\!-\!1\,$ then $\, {\rm mod}\ p\!:\ a^e\equiv a.\,$ Indeed, it is true if $\,a\equiv 0,\,$ and if $\,a\not\equiv 0\,$ then $\,e\, =\, 1+k(p\!-\!1)\ $ so $\ a^e\equiv a (\color{#c00}{a^{p-1}})^k\overset{\rm\color{#c00}{ Fermat}}\equiv a \color{#c00}{(1)}^k\equiv a,\ $ so $\ a^e\!-a\equiv a-a\equiv 0.$

In your case we have $\,2\!-\!1,\,3\!-\!1,\,7\!-\!1$ all divide $\,e\!-\!1 = 7\!-\!1,\,$ so $\,a^7-a\,$ is divisible by $\,2,3,7,\,$ therefore it is also divisible by their lcm = product $ = 42.$

Remark $\ $ Above we used basic congruence rules. The key idea generalizes, see Korselt's Carmichael Number Criterion.

Community
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Bill Dubuque
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  • This looks great but it's way beyond my reach congrats though you are very much a genius. – James K Mar 06 '15 at 22:06
  • @James You can safely ignore the Remark. Everything before it is very easy, using only Fermat's little Theorem and congruences. If anything is not clear, please feel welcome to ask questions, and I will be happy to explain further. – Bill Dubuque Mar 06 '15 at 22:09
  • No worries mate. I know that it is just me now. Definitely starting to think I am not cut out for this as it takes me ages to understand things and I can never get answers on my own. Appreciate the effort though :) – James K Mar 06 '15 at 22:11
  • @James Do you know Fermat's little Theorem, and congruence (modular) arothmetic? If so, then you can understand this. Please tell me where I lost you and I can elaborate. – Bill Dubuque Mar 06 '15 at 22:13
  • Yes I know FLT but I never understood the proof when I saw it in class. I think I understand this question now thanks. – James K Mar 06 '15 at 22:16
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In fact, $42$ is the largest integer that divides $a^7-a$ for all $a\in\Bbb Z$.

$a^7-a=a\left(a^6-1\right)=a\left(a^3-1\right)\left(a^3+1\right)$

$=a(a-1)\left(a^2+a+1\right)(a+1)\left(a^2-a+1\right)$

$a-1,a,a+1$ are three consecutive integers, so $2,3\mid a^7-a$.

Also by Fermat's Little theorem $7\mid a^7-a$. Therefore $2\cdot 3\cdot 7=42\mid a^7-a$.

$2^7-2=42\cdot 3$ and $3^7-3=42\cdot 52$. But $\gcd(3,52)=1$,

so it follows that $42$ is the largest integer that divides $a^7-a$ for all $a\in\Bbb Z$.

user236182
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Use Fermat's little theorem mod 2, 3 and 7 since these are the prime factors of 42.

Dean Barber
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First by the Fermat's little theorem we have $$a^7\equiv a \bmod 7$$ so $7|a^7-a$. On the other hand by applying the same theorem it is easy to see that $$2|a^7-a$$ and $$3|a^7-a \quad*$$

Proof of $3|a^7-a$.

If a is a multiple of $3$, * is clear. So suppose that $\gcd (a,3)=1$. Then Fermat's theorem implies that $$a^2\equiv a\bmod 3\implies a^6\equiv a^3\equiv a^2\times a\equiv a\times a\equiv a \bmod 3 \implies a^7\equiv a^2 \equiv a\bmod 3$$ Could you repeat this for the other case?

Fermat
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