Corollary to Euler's Theorem

When we studied Euler's Theorem back in college (in Switzerland), the professor mentioned the following simple corollary, leaving the proof as an exercise: $a^{\phi(n)+k}\equiv{a^k}\pmod{n}$ for all positive integers $a$ and $n$ , as long as $k$ is $\geq$ the multiplicity of all the primes in the factorization of $n$ . Can you prove (or disprove) this?

For example, if $n=2^3\times{5}\times{7^4}$ , we want $k\geq{4}$

Thus we can say that "in modular arithmetic, all exponential functions eventually become periodic."

#Algebra

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Comments

Let me try to outline a proof.

It suffices to prove the congruency modulo all the prime power factors of $n$ . For example, if $n=2^3\times{5}\times{7^4}$ , it suffices to prove the congruency modulo $2^3, 5$ and $7^4$ .

So, let $p^m$ be a prime power factor of $n$ . If $p\not|{a}$ , then we have $a^{\phi(p^m)}\equiv{1}\pmod{p^m}$ by Euler's Theorem. Since $\phi(p^m)|\phi(n)$ , we have $a^{\phi(n)}\equiv{1}\pmod{p^m}$ as well and therefore $a^{\phi(n)+k}\equiv{a^k}\pmod{p^m}$ as claimed.

If $p|a$ , then $p^m|a^k$ since $k\geq{m}$ , so that $a^{\phi(n)+k}\equiv{0}\equiv{a^k}\pmod{p^m}.$

Otto Bretscher - 6 years ago

Can't we use euler's theorem and multiply a^k to both sides giving result.

shivamani patil - 6 years ago

Euler's theorem is valid iff $\gcd(a,n)=1$ where $a,n\in\Bbb{Z^+}$ . The result in this note, however, doesn't have the restriction that $\gcd(a,n)=1$ and hence you cannot directly use Euler's Theorem here.

Prasun Biswas - 6 years ago

@Prasun Biswas – Forgot it .Now it seems slight difficult thanx.

shivamani patil - 6 years ago

@Prasun Biswas – Ohk , thanx!

Harsh Shrivastava - 6 years ago

@Prasun Biswas – Ya here we have to deal extra case of gcd not being 1 rest seems using euler theorem and some arithmetic .

shivamani patil - 6 years ago

Yes i also thought the same.....

Harsh Shrivastava - 6 years ago

@Tijmen Veltman proved this for $k=1$ and square-free $n$ in this note.

He might be interested in proving/disproving this one too, so I'm tagging him here.

Prasun Biswas - 6 years ago

Interesting! Thank you for letting me know.

The result I quote is sometimes mentioned in introductory Number Theory texts, usually as an exercise. I have never found any use for it... until I joined Brilliant ;)

Otto Bretscher - 6 years ago

Let the GCD of $a$ and $n$ be $g$ . Then, $a=gc$ and $n=gd$ .

The rest of the proof is left to the reader as an exercise. (Just kidding, I got to go now and I will finish it later.)

Kenny Lau - 5 years, 8 months ago

I made a problem about this a while back (although I just now read this note): Divisibility of power differences. It was when I was writing the wiki on Carmichael numbers.

Patrick Corn - 5 years ago

Proof for the corollary:

let $n = p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{m}^{a_m}$ be the standard factorisation of $n$ .

$k$ is greater than the multiplicity of all the primes in the factorization of $n$ is equivalent to $k \geq max\left\{a_1, a_2, ..., a_m\right\}$ .

if $a, p_i$ are co-prime, $a^{\phi(n) + k} = a^{\phi(p_{1}^{a_{1}})\phi(p_{2}^{a_{2}})...\phi(p_{m}^{a_m}) + k} \equiv a^k \pmod{p_{i}^{a_i}}$ .

if $a, p_i$ are not co-prime, $v_{p_i}(a^k) = kv_{p_i}(a) \geq a_i \Rightarrow a^k \equiv 0 \pmod{p_{i}^{a_i}}$ , where $v_{p}(n)$ denotes the exponent of $p$ in the prime factorization of $n$ . $\phi(n) + k > k \Rightarrow a^{\phi(n) + k} \equiv 0 \pmod{p_{i}^{a_i}}$ . Hence $a^{\phi(n) + k} \equiv a^k \equiv 0 \pmod{p_{i}^{a_i}}$ .

Hence, for all positive $a, n$ , $a^{\phi(n) + k} \equiv a^k \pmod{p_{i}^{a_i}}$ .

Since this is true for all $i \in \left\{1, 2, ..., m\right\}$ , the result follows.

George Edgar - 1 year, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$