Nontotient numbers
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Started by: chiph588chiph588
On: 1228079953|%e %b %Y, %H:%M %Z|agohover
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Nontotient numbers
chiph588chiph588 1228079953|%e %b %Y, %H:%M %Z|agohover

A nontotient number n \in \mathbb{N} is a number such that there is no x \in \mathbb{N} where \phi(x) = n.

The smallest such number is 14.

My question is how would one prove for example that 14 is nontotient?

unfold Nontotient numbers by chiph588chiph588, 1228079953|%e %b %Y, %H:%M %Z|agohover
Re: Nontotient numbers
jensbergjensberg 1228108345|%e %b %Y, %H:%M %Z|agohover

I think that it's not that bad. But you also left out the fact that all odd numbers (excluding 1) are nontotient.
So remembering that \phi(x) is multiplicative, we note the following: suppose n = p_1^{a_1} * p_2^{a_2}, with p_1, p_2 \neq 2 (there could be a 2 in there, but not more than one, since \phi(2) = 1 but that 4|\phi(2^k) and 14 isn't divisible by 4). Then \phi(n) = p_1^{a_1 -1}(p_1 -1) + p_2^{a_2 -1}(p_2 -1). But 2|p_1 -1 and 2|p_2 -1 and thus that implies that 4|\phi(n) which would be impossible. And thus n cannot be a product of odd primes. So the only case left is that n=p^{a} (again possibly times 2), and since p is odd let p = 2k+1, k \in \mathbb{Z}. But then \phi(n) = p^{a-1}(p-1) which implies that \frac{14}{2} = 7 = (2k+1)^{a-1}*k. And then there's no k that makes that work (that would need to be more rigorous, but it's certainly a start).

unfold Re: Nontotient numbers by jensbergjensberg, 1228108345|%e %b %Y, %H:%M %Z|agohover
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