For what integers does phi(n) | n and why?

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- January 28th 2009, 04:32 AMCairoEuler Totient
For what integers does phi(n) | n and why?

- January 28th 2009, 03:25 PMclic-clac
Let be a prime and a positive integer, .

So you see that the only prime such that is .

Let be an integer such that , and its unique decomposition in a product of primes, with and . Then, for all has to divide , and a consequence is .

so , and since is odd, that means .

If , that's ok; if , and can't divide and

As you may see, .

Furthermore, .

Conclusion: the integers such that are the elements of - January 29th 2009, 09:08 AMCairo
Thank you for this.

I'm a little confused over your notation, since you appear to interchange the exponent k and l for your primes.

I have managed to google an exam paper that suggests that the answer is n = 2^a*3^b for a, b in N.

How does this work? - January 29th 2009, 10:46 PMclic-clac
Hi. The exponent notations can be changed during the proof, they are unknowns, so there is no problem.

But the integers arn't the solution because doesn't divide (that's why the solution is a little bit more complicated)