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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."

Note by Otto Bretscher
1 year, 8 months ago

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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 · 1 year, 8 months ago

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@Otto Bretscher Can't we use euler's theorem and multiply a^k to both sides giving result. Shivamani Patil · 1 year, 8 months ago

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@Shivamani Patil 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 · 1 year, 8 months ago

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@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 · 1 year, 8 months ago

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@Prasun Biswas Ohk , thanx! Harsh Shrivastava · 1 year, 8 months ago

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@Prasun Biswas Forgot it .Now it seems slight difficult thanx. Shivamani Patil · 1 year, 8 months ago

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@Shivamani Patil Yes i also thought the same..... Harsh Shrivastava · 1 year, 8 months ago

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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 · 7 months, 3 weeks ago

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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, 3 months ago

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@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 · 1 year, 8 months ago

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@Prasun Biswas 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 · 1 year, 8 months ago

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