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: aϕ(n)+kak(modn)a^{\phi(n)+k}\equiv{a^k}\pmod{n} for all positive integers aa and nn, as long as kk is \geq the multiplicity of all the primes in the factorization of nn. Can you prove (or disprove) this?

For example, if n=23×5×74n=2^3\times{5}\times{7^4} , we want k4k\geq{4}

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

Note by Otto Bretscher
4 years, 5 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 nn. For example, if n=23×5×74n=2^3\times{5}\times{7^4} , it suffices to prove the congruency modulo 23,52^3, 5 and 747^4.

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

If pap|a, then pmakp^m|a^k since kmk\geq{m} , so that aϕ(n)+k0ak(modpm).a^{\phi(n)+k}\equiv{0}\equiv{a^k}\pmod{p^m}.

Otto Bretscher - 4 years, 5 months ago

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

shivamani patil - 4 years, 5 months ago

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

Prasun Biswas - 4 years, 5 months ago

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

shivamani patil - 4 years, 5 months ago

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@Prasun Biswas Ohk , thanx!

Harsh Shrivastava - 4 years, 5 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 - 4 years, 5 months ago

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Yes i also thought the same.....

Harsh Shrivastava - 4 years, 5 months ago

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

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

Prasun Biswas - 4 years, 5 months ago

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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 - 4 years, 5 months ago

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Let the GCD of aa and nn be gg. Then, a=gca=gc and n=gdn=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 - 4 years 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 - 3 years, 5 months ago

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