Proving the reccuring pattern of $60^n\equiv1,4,2 \mod 7$

Question

By simple trial and error I stumbled upon:
$60^0\equiv1 \mod 7$
$60^1\equiv4\mod 7$
$60^2\equiv2 \mod 7$
$60^3\equiv1 \mod 7$
$60^4\equiv4\mod 7$
$60^5\equiv2 \mod 7$
...

This goes on forever. Well maybe, that's the thing I have no idea how you would prove this.

I suppose induction is possible but what are the base cases here?

$\bmod 7!:\ 60\equiv 2^2,$ so $,60^{\bf \large \color{#c00}3}\equiv 2^6\equiv 1,$ by the Congruence Power Rule and Fermat's little Theorem, so we deduce $,60^{\large n}\equiv 60^{\bf\large n\bmod\color{#c00} 3},$ by mod order reduction in the linked dupe. — Bill Dubuque, Mar 22 '22 at 15:51

score 0 · Accepted Answer · answered Mar 22 '22 at 14:34

First of all $60\equiv4\pmod 7$ as $56$ is divisible by $7$.

Now if you actually know Fermat's Little Theorem then you would see that for any integer $a$ coprime to a prime number $p$. You have $a^{p-1}\equiv 1\pmod p$.

So $4^{7}\equiv 4^{6}\cdot 4\pmod 7\equiv 1\cdot 4\pmod 7 \equiv4\pmod 7$ and so the pattern continues and after each multiple of $6$ you will get back the same integer. So it suffices to show that the first $6$ powers are like that. Then the pattern holds for all integers .

Maybe you can take a look at some basic modular arithmetic to make things clearer. Maybe someday you will also learn group theory and cyclic groups. Maybe someday you will get to learn about Euler's Totient function or Carmichael's Lambda Function

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. — Bill Dubuque, Mar 22 '22 at 15:51

Proving the reccuring pattern of $60^n\equiv1,4,2 \mod 7$

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