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Math Help - Euler's Totient function

  1. #1
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    Euler's Totient function

    The problem is to show that there are infinitely many odd integers k for which φ(k) is some number squared.

    Now, I know that there are an infinite amount of positive integers that would make φ(k) a squared number .
    Taking k=22m+1, where m is a non-negative integer:
    φ(22m+1)=22m+1-22m=22m(2-1)=(2m)2

    But I'm having trouble finding how to show that this is true for odd k.

    I wrote out a list for φ of odd numbers, and found that this works for k = 5, 17, 37, 57, 63, 85, 185, 629, 2255, but I don't see a pattern emerging.

    I know that IF there were an infinite amount of primes of the form n2+1, then I could take k=n2+1 and the result would immediately follow since φ(p)=p-1, where p is a prime. But I'm almost certain I'm not allowed to use that conjecture. :P

    Does anyone have some suggestions as to what else I should try?

    Thanks for your time!
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  2. #2
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    Re: Euler's Totient function

    Consider k = 5m17n for suitable choices of m and n.

    ETA: On second thought, just consider k = 5m for suitable m.
    Last edited by Petek; April 18th 2012 at 01:53 PM.
    Thanks from mi986
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  3. #3
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    Re: Euler's Totient function

    Oooh!
    So, it's just like the case with 2!
    take k=52m+1, where m is a non-negative int. Then,
    φ(52m+1)
    =52m+1-52m
    =52m(5-1)
    =(5m2)2

    Ah, this means I can get similar results for 17, 37, 101, etc...
    That's actually kind of awesome.

    Thanks so much.
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