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Totient Function and phi(n)=phi(2n) for odd n

  1. Jul 6, 2008 #1

    Last year in number theory I discovered something with Euler's Totient Function that I couldn't explain. I asked my professor and he couldn't figure it out either. Here is what I discovered:

    As a premise, it should be clear that phi(n)==phi(2n) where n is an odd, positive, integer. Using the formula for Euler's totient function, this is clear. However, looking at the numbers, it becomes more complex. For example, let n = 9. We know that phi(9)=phi(18)=6.

    So we have the pos. integers from 1 to 9 ( n in red, coprime integers in blue, noncoprime integers in orange). Note that there are 4 primes in this range (2,3,5,7):

    [itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{blue}2}, {\color{orange}3}, {\color{blue}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{blue}8}, {\color{red}9} [/itex]

    Next, the integers from 1 to 18 for n=18. Note there are only 3 new primes from 9-18

    [itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{orange}2}, {\color{orange}3}, {\color{orange}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{orange}8}, {\color{orange}9}, {\color{orange}10}, \color{blue}11}, \color{orange}12}, \color{blue}13}, \color{orange}14}, \color{orange}15}, \color{orange}16}, \color{blue}17}, \color{red}18}[/itex]

    I'm having a hard time phrasing this, but I think what I mean to say is this: How can the phi function know to compensate for the number of prime numbers between n and 2n? Wouldn't this be related to the density of primes? Also, could this be used to prove that there exist infinite prime numbers?

    I'm sorry If my wording is hard to understand. Once I gain some more understanding, I would like to write it up properly and show it to my math teacher or something.

    Thanks in advance.
  2. jcsd
  3. Jul 9, 2008 #2
    That's interesting. I'll try to get something to get back to you with.
  4. Jul 9, 2008 #3
    This is no more than a comment. It is not an answer. Hence best phrased as a question.

    You ask "How can the phi function know to compensate for the number of prime numbers between n and 2n?" But, does the data indicate that it really does that as opposed to compensating for the number of co-prime numbers between n and 2n?

    Nothing more than a (hopefully) seminal question, but, I really would like it if you could give me your best shot at an answer.

  5. Jul 9, 2008 #4
    Upon further investigation, it seems as though you are right. Here is what I came up with (please forgive my lack of precision in stating x is a positive integer, etc.):

    \phi (2n)=\left{|}\{x|x\leq 2n-1\}-\{x|x \textrm{ is even}\} - \{i\cdot x | x \textrm{ odd}, x|n, i \textrm{ odd}\}\right{|}

    So, for [itex]n=15[/itex], we have:

    \phi (30) = |\{1,\dots, 29\}-\{\textrm{evens}\}-\{3,9,15,21,27,5,25\}|
    \phi (30) = | 29-14-7 | = 8

    As we already know, [itex]\phi (30)=\phi (15)=8[/itex], and thus this method works. So it would seem that it has nothing to do with the number of primes between 1 and 2n.

    Please give me your thoughts.
  6. Feb 20, 2010 #5
    The formula for euler's function ϕ(n) is given by

    ϕ(n) = n (1-1/p1) (1 – 1/p2) … (1 – 1/pk)
    where n is prime factorized into p1a1p2a2...pkak and each pk is a prime.

    Assume n is an odd number. None of the pk will be an even number. When n is multiplied by 2, the prime factorization of n will b (p1^a1)(p2^a2)...(pk^ak) * 2. The formula for ϕ(n) for 2n will be

    ϕ(2n) = 2n ((1-1/p1) (1 – 1/p2) … (1 – 1/pk) (1 - ½) = n ((1-1/p1) (1 – 1/p2) … (1 – 1/pk)
    which equals ϕ(n)

    Assume n is an even number. Multiplying n by 2 will increase the power of 2 by 1.

    ϕ(2n) = 2n ((1-1/p1) (1 – 1/p2) … (1 – 1/pk) = 2ϕ(n)

    You're welcome.
  7. Feb 20, 2010 #6
    n=17 could be another interesting example. In the upper half of the coprimes to 34, you find not only new primes (19,23,29,31), but also powers of smaller primes (25,27) as well as other composite numbers (21,33).

    As n gets bigger, the panorama gets only more varied.
  8. Feb 20, 2010 #7
    Doesn't ϕ(mn) = ϕ(m)ϕ(n)?

    And doesn't ϕ(2) = 1?

    What's the mystery?
  9. Feb 21, 2010 #8

    ϕ(2m) = 2ϕ(m) ONLY IF m is even thats where's the mystery. And i think I did a pretty good job explaining it in post #5 if anyone is interested...
  10. Feb 21, 2010 #9
    But what does "even" mean? It means no factors of two.

    Have you noticed that ϕ(3m) is twice ϕ(m) if m is not divisible by 3?

  11. Feb 21, 2010 #10
    Are you serious?
  12. Feb 21, 2010 #11
    If the OP's question is why phi(2n) = phi(n) when n is odd, I think my explanation is clear enough,.... see post #5.

    please feel free to ask questions and dont say stuff like even numbers have no factor of two.
  13. Feb 21, 2010 #12
    Of course not. That was a brain fart.
  14. Feb 22, 2010 #13
  15. Feb 22, 2010 #14


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    If there were finitely many primes, then phi(n) >= kn for some fixed positive k. If you know that there is an infinite sequence a_n with phi(a_n) < k a_n/log log a_n for some fixed positive k, then you could use this to show that there are infinitely many primes. But that's proving an easy result with a hard result.
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