## Empty one year ago Prove the following:

1. Empty

If n>1 and $$\phi(n)$$ is Euler's totient function, then $\phi(n) \equiv 1 \mod 2$ implies that $$n=2$$

2. ganeshie8

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

that proves the contrapositive : $$n\ne 2 \implies \phi(n)\not\equiv 1\pmod{2}$$

4. Empty

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

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

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

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

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

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

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

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

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

14. ganeshie8

just need to notice that $$\phi(n)$$ represents the number of positive integers less than $$n$$ such that $$\gcd(k,n)=1$$

15. Empty

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

|dw:1439268673998:dw|

17. ganeshie8

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

from the diagram, notice that the cardinalities of subsets add up to the cardinality of the set

19. ganeshie8

|dw:1439269268456:dw|

20. Empty

Hmmm I don't quite follow, I'll sleep on it and read this tomorrow I think I just need some rest.