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  • one year ago

Prove the following:

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  1. Empty
    • one year ago
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    If n>1 and \(\phi(n)\) is Euler's totient function, then \[\phi(n) \equiv 1 \mod 2\] implies that \(n=2\)

  2. ganeshie8
    • one year ago
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    Consider two cases : 1) If \(n\) has a factor of odd prime, \(p\), then \[\phi(n)=\phi(p^km)=\phi(p^k)\phi(m)=(p-1)(p^k-1)\phi(m)\equiv 0\pmod{2}\] 2) if \(n\) has no odd prime factors, then \[\phi(n)=\phi(2^k)=2^k-2^{k-1}=2^{k-1}\equiv 0\pmod{2} ~~\text{for k} \gt 1\]

  3. ganeshie8
    • one year ago
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    that proves the contrapositive : \(n\ne 2 \implies \phi(n)\not\equiv 1\pmod{2}\)

  4. Empty
    • one year ago
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    Interesting! I had come up with this problem through a slightly different way so it was nice to see that this could be proven this way! Thanks!

  5. ganeshie8
    • one year ago
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    May I know your different way... There is an interesting alternative proof using below property of \(\phi\) function : \[\large \sum\limits_{1\le k\lt n, ~\gcd(n,k)=1}~k~ = ~\frac{1}{2}n\phi(n)\]

  6. Empty
    • one year ago
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    Yeah, essentially that's what I understand but maybe I found it less formally. From the thing I posted in ikram's question the other day, I realized that since relatively prime numbers come in pairs that would imply \(\phi(n)\) is divisible by 2 for all values greater than 2.

  7. ganeshie8
    • one year ago
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    Exactly! that is really a cool observation Since \(\gcd(k,n)=1\iff \gcd(n-k, n)=1\), below two sets are identically equal : \[\{k_1,~k_2,~\ldots,k_{\phi(n)}\}\] and \[\{n-k_1,~n-k_2,~\ldots,n-k_{\phi(n)}\}\] Easy to see that each \(k_i\) pairs up with \(n-k_i\)

  8. Empty
    • one year ago
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    Yeah, now if I can show that at least one of these pairs contains both prime numbers, then you've proven the goldbach conjecture haha. ;P

  9. Empty
    • one year ago
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    Also while playing with the \(\phi\) function I realized a fun little way to easily remember \[n = \sum_{d|n} \phi(d)\] Let's just look at one part of it (after all we know \(n\) is multiplicative) \[\sum_{i=0}^{k} \phi(p^i) \] plugging in \[\phi(p^i) = p^i - p^{i-1}\] Well that's just going to be a telescoping series, so clearly \[p^k = \sum_{i=0}^{k} \phi(p^i) \] Anyways in some sense this almost reminds me of like the fundamental theorem of calculus or Green's theorem, but anyways so something kinda interesting to share and relate.

  10. Empty
    • one year ago
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    Err ok I fixed it, but yeah by that I mean that \(\phi(n)\) is like the derivative of n and the sum over the divisors of n is like the integral: \[n = \sum_{d|n} \phi(d)\] So sorta funny way to look at it maybe. I think I am really just satisfied by the fact that its' a telescoping series honestly. Also we of course might say that if this is really the fundamental theorem of calculus in the "dirichlet sense" whatever this means we also have an alternate way to compute the "derivative" \[\phi(n) = n * \mu(n)\]

  11. ganeshie8
    • one year ago
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    That is actually one of the proofs of \(n = \sum\limits_{d\mid n}\phi(d)\), but I like more the one which uses just the divisibility argument. Consider the following partition of integers between \(1\) and \(n\) : \[S_d=\{m~|~\gcd(m,n)=d; ~1\le m\le n\}\] Since \(\gcd(m,n)=d \iff \gcd(m/d,n/d)=1\), we have \(|S_d|=\phi(n/d)\). So \[\sum\limits_{d\mid n} |S_d| = \sum\limits_{d\mid n} \phi(n/d) = \sum\limits_{d\mid n} \phi(d)\] the left most sum equals \(n\) because those gcd subsets forma a partition.

  12. Empty
    • one year ago
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    I saw that on MSE but I actually don't fully understand this argument for some reason. I guess it's when they say \(|S_d| = \phi(n/d)\) about there that I lose it I think.

  13. ganeshie8
    • one year ago
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    can you provide the link

  14. ganeshie8
    • one year ago
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    just need to notice that \(\phi(n)\) represents the number of positive integers less than \(n\) such that \(\gcd(k,n)=1\)

  15. Empty
    • one year ago
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    I can't seem to find it but they used the exact same argument you did and even called the set \(S_d\) like you did too. Oh well I think I am starting to understand it now, what I don't understand is what "forming a partition" means.

  16. ganeshie8
    • one year ago
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    |dw:1439268673998:dw|

  17. ganeshie8
    • one year ago
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    Easy to see that the gcd subsets form a partition :- they are disjoint because each of the integers from \(1\) to \(n\) can be member of exactly one set (\(S_d\)) as gcd maps each integer to another unique integer (function property)

  18. ganeshie8
    • one year ago
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    from the diagram, notice that the cardinalities of subsets add up to the cardinality of the set

  19. ganeshie8
    • one year ago
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    |dw:1439269268456:dw|

  20. Empty
    • one year ago
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    Hmmm I don't quite follow, I'll sleep on it and read this tomorrow I think I just need some rest.

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