While experimenting with modulo arithmetic I came upon this slight extension to Fermat's Little Theorem (for which I have a proof).

$$\text{If a is a positive integer and p is prime and }\ge\text{ 5 then}$$

$$a^p \equiv a\pmod{3\times p} $$

I cannot believe this has remained unnoticed but I will ask anyway.

Is there any chance this is new?

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    $\begingroup$ Is the proof too large to fit in the margins of the question box? $\endgroup$ – Oscar Lanzi Oct 5 '17 at 14:26
  • $\begingroup$ Actually, $a^p\equiv a\pmod{6p}$. $\endgroup$ – Thomas Andrews Oct 5 '17 at 15:12
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    $\begingroup$ More generally, if $p-1$ is divisible by $q-1$ for $p\neq q$ prime, then $a^p\equiv a\pmod{qp}$ $\endgroup$ – Thomas Andrews Oct 5 '17 at 15:20
  • $\begingroup$ Very special case of this generalized Fermat Euler Theorem. $\endgroup$ – Bill Dubuque Oct 10 '18 at 21:27

This is certainly known, but it doesn't have a name.

More generally,

Lemma: If $q$ is prime and $n$ is any positive integer such that $q-1\mid n-1$ then: $$a^n\equiv a\pmod{q}$$ Proof: This can be proved by induction on $k=\frac{n-1}{q-1}$.

Let $n_k=1+(q-1)k$. Then $n_0=1$ and $a^{1}\equiv a\pmod{q}$.

Now, if $a^{n_k}\equiv a\pmod{q}$, then $$a^{n_{k+1}}=a^{n_k}a^{q-1}\equiv a\cdot a^{q-1}=a^q\equiv a\pmod{q}$$

Now the case when $p\geq 5$ is prime and $q=3$ means $a^{p}-a$ is divisible by $p$ and $3$, thus, by $3p$.

It is also divisible by $2$, so you really get:

$$a^{p}\equiv a\pmod{6p}$$

You also have that if $p>5$ with $p\equiv 1\pmod 4$ then $a^p\equiv a\pmod{30p}$.

The lemma lets you assert, for example, that $a^{13}\equiv a\pmod{2\cdot 3\cdot 5\cdot 7\cdot 13}$.

  • $\begingroup$ Thanks to everyone who replied, there is plenty for me to chew on here! $\endgroup$ – Robert G. Burns Oct 5 '17 at 15:36

Try these problem by yourself first: 1. Given relatively prime integers $a, b, c$, if I tell you any three numbers $i, j, k$, can you tell me an integer $N$ such that all three holds simontaneously: $$ N \equiv i \mod a$$ $$N \equiv j \mod b$$ $$N \equiv k \mod c$$ ? Can you always give me such $N$ ?

As a concrete example, can you tell me a number $N$ such that $$N \equiv 5 \mod 7$$ $$N \equiv 2 \mod 5$$ $$N \equiv 10 \mod 11$$ ?

Can you generalize ?

  1. When $N$ varies from $0$ to $5*7*11$, how does $$\{ N \mod 7 \}, \{ N \mod 5 \}, \{N \mod 11 \}$$ varies ?

Nice observation, but this is basically Chinese Remainder Theorem. (See the handout here for some nice and hard problems regarding this)

The $\mod p$ part is handled by Fermat's theorem, but the $\mod 3$ part is easily handled by observing $$ 2^{odd} \equiv 2 \mod 3$$ and $$1^{odd} \equiv 1 \mod 3$$.

In case you're feeling bad this turned out to be trivial, try this fun one if you didn't see it before: $2n$ divideds $\phi(a^n+1)$ for any integer $a, n$ with $gcd(a,n) = 1$

  • $\begingroup$ I am sorry but don't understand why this is the Chinese Remainder Theorem (i'm not saying it isn't). Can I give an example in case my statement was imprecise? Take p dto be 7 and a to be 46, then 46 mod(21) is 4. Also 46^7 is 435817657216 and 435817657216 is 4 mod(21). And this is true for any a. $\endgroup$ – Robert G. Burns Oct 5 '17 at 14:51
  • $\begingroup$ @RobertG.Burns No problem $: )$. I have put two problems for you - try it by yourself and tell me the solutions in the comment after you've figured it out. If you still can't find the connection after thinking hardly, only then I will tell you $: )$ $\endgroup$ – cdt Oct 5 '17 at 14:54

Actually by Fermat's little theorem, $3$ always divides $$a^3-a=a(a^2-1)$$

As $p-1$ is even for $p\ge3$ $$a^{p-1}-1$$ is divisible by $a^2-1$

Hence, $a^p-a$ will be divisible by $3$ and $p$ and consequently by lcm$(3,p)$

  • $\begingroup$ So it's true but obvious? $\endgroup$ – Robert G. Burns Oct 5 '17 at 15:14
  • $\begingroup$ @RobertG.Burns, u r right $\endgroup$ – lab bhattacharjee Oct 5 '17 at 16:04

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