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."

## Comments

Sort by:

TopNewestLet 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 · 2 years ago

Log in to reply

– Shivamani Patil · 2 years ago

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

– Prasun Biswas · 2 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.Log in to reply

– Shivamani Patil · 2 years ago

Ya here we have to deal extra case of gcd not being 1 rest seems using euler theorem and some arithmetic .Log in to reply

– Harsh Shrivastava · 2 years ago

Ohk , thanx!Log in to reply

– Shivamani Patil · 2 years ago

Forgot it .Now it seems slight difficult thanx.Log in to reply

– Harsh Shrivastava · 2 years ago

Yes i also thought the same.....Log in to reply

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 · 11 months, 3 weeks ago

Log in to reply

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 · 1 year, 7 months ago

Log in to reply

@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 · 2 years ago

Log in to reply

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 · 2 years ago

Log in to reply