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I have not broken this down very far. I have come to the conclusion that there are infinitely many values for n where there exists 12 coprimes to n. Since there are infinitely many primes, and primes are coprimes to any number smaller than that prime, I reach that conclusion. Can anyone stop me and tell me where I'm going wrong or how to approach this.

I've used the theorem:

$ \varphi (n) = ((1-\frac{1}{p_{1}})...(1-\frac{1}{p_{k}})) $

But to compute these values would one not need to write a program to run which slots in all these primes?

Gregory Peck
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    There are only finitely many positive integers whose totient is a given positive integer. – Nishant Oct 22 '14 at 21:13
  • Could you expand on that, and point me in the right direction. Sorry if I seem a bit ignorant on the subject – Gregory Peck Oct 22 '14 at 21:16
  • Finiteness also proved in my answer here – ccorn Oct 22 '14 at 21:17
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    No, it's not a duplicate. That other OP was not laboring under false conclusions. – James47 Oct 22 '14 at 21:24
  • If you can write a program to compute this, then you can do it by hand (just do what you propose to tell the computer to do). This might give you more intuition about what is going on. – Cardboard Box Oct 22 '14 at 21:37
  • I think marking this a duplicate was a boneheaded decision. Maybe the two askers were motivated by the very same textbook, but this asker went further, albeit in the wrong direction. – Robert Soupe Oct 23 '14 at 02:36

3 Answers3

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Elaborating on James47's answer as he requested: Write $n = q_1 \cdots q_k$ for $q_i = p_i^{n_i}$ with the $p_i$ distinct. Then $$\phi(n) = \phi(q_1) \cdots \phi(q_k) = p_1^{n_i - 1} \cdots p_k^{n_k - 1}(p_1 - 1) \cdots (p_k - 1) = 12.$$ It's then just a matter of listing the $(q_1, \dots, q_k)$ satisfying the equation above. In particular, each $p_i^{n_i - 1}$ and $p_i - 1$ must divide $12$, which limits the possibilities to $q_i = 2, 2^2, 2^3, 3, 3^2, 5, 7,$ and $13$. That should be enough to brute-force the result.

anomaly
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Your conclusion is wrong. For now, forget the formula involving reciprocals.

If $p$ is prime, then $\phi(p) = p - 1$. Given $k > 1$, we have $\phi(p^k) = p^{k - 1}(p - 1)$. Also, this function is multiplicative, meaning that $\phi(pq) = \phi(p) \phi(q)$ if $\gcd(p, q) = 1$. Furthermore, there are only two exceptions to $\phi(m) > \sqrt{m}$.

So, to find all solutions for $\phi(n) = 12$ you need to figure out the finite set of ways to express $12$ as a product of numbers $1$ less than a prime and/or powers of a prime times $1$ less than that prime.

  • $12$ is one less than prime, $13$.
  • $12 = 2 \times 6$, and $21 = 3 \times 7$.
  • $12 = 1 \times 12$. This might seem obvious and pointless, until you look at $26 = 2 \times 13$
  • $28 = 2^2 \times 7$ corresponds to $12 = 2 \times 6$ because $\varphi(4) = 2$
  • $36 = 2^2 \times 3^2$ also corresponds to $12 = 2 \times 6$ because $\varphi(9) = 6$
  • $42 = 2 \times 3 \times 7$ also corresponds to $12 = 2 \times 6$

and you don't need to look at oh dang, I don't have time to finish up this answer anyone else do me the favor?

EDIT: Robert Soupe here. I completed the table explaining the correspondence between values of $n$ such that $\varphi(n) = 12$ and products that give 12. I really don't know where Jim is going with that stuff about $\varphi(n) > \sqrt{n}$ (though I do know the two exceptions: 2 and 6).

Robert Soupe
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James47
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First of all, be careful, you forgot a term in your expression of $\varphi{(n)} = n \left(1-\dfrac{1}{p_{1}}\right)...\left(1-\dfrac{1}{p_{k}}\right)$ : there is a $n$.

Developping the first formula, if $n = p_1^{\alpha_1}...p_n^{\alpha_n}$, then $\varphi{(n)} = p_1^{\alpha_1-1}(p_1-1)...p_n^{\alpha_n-1}(p_n-1).$

If $\varphi{(n)} = 12 = 2*2*3$, we only have to find all the possibilities for the $p_i$ and $\alpha_i$.

If $3$ is alone, it is a $p_i^{\alpha_i-1}$ term (otherwise it would be a $(p_i-1)$ term, and $p_i$ would be $4$, which isn't prime). It goes inevitably with a $(p_i-1)$ term of $2$.

We only have a $2$ left. Hence the only possible match for him is a $(2*1)$ since a $(1*2)$ would imply another $p_i = 3$. This yields $n = 36$ and is the unique possibility with $3$ not being paired.

$3$ can be paired with a $2$ to form a $6$. This $6$ is obviously a $(p_i-1)$ term, so $p_i$ is $7$, with multiplicity $1$ because $7$ is not in our list of factors. The last $2$ can exist in the form $(2*1)$ or $(1*2)$. This yields $n = 28$ and $21$ respectively. Let's not forget the possible $(1*1)$ that can happen in the second case since $2$ isn't a factor in this case. This gives us $n=42$.

Finally, there can exist a lonely $3*2*2$ term. Once again since $12$ is not prime, it is a $(p_i-1)$ term, with $p_i$ being $13$ with multiplicity $1$. But this is not over yet. Even if we have no factors left, we can still have a $(1*1)$ term, like before. If it is there, then $n = 26$. Otherwise, $n=13$.

The final list is : $13, 21, 26, 28, 36, 42$.

Traklon
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