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I just recently started working with Euler's Totient Function, and I came across the problem of solving for all possible integers $n$ such that $\varphi(n)=14$. I know there are similar questions with different numbers, I just wanted to verify if my approach was correct or I messed up since I could not find any $n$ satisfying that property.

We know $$\varphi(n)= \prod_{i=0}^{k} p_{i}^{a-1}(p_i-1)$$ From there we see $p_i \leq 15$ hence the primes are $p_i=\{2, 3, 5, 7, 11, 13 \}$. Ergo, we can write $n= 2^a\cdot3^b\cdot5^c\cdot7^d\cdot11^e\cdot13^f$. Using the fact that the Function is multiplicative, we can separate $$\varphi(2^a\cdot\ldots\cdot13^f)= \varphi(2^a)\cdot\ldots\cdot \varphi(13^f) = 2^{a-1}(2-1)\cdot3^{b-1}(3-1)\cdot\ldots\cdot13^{f-1}(13-1) $$

So I ended up with $\varphi(n)= 2^{a+6}\cdot3^{b+1}\cdot5^{c}\cdot7^{d-1}\cdot11^{e-1}\cdot13^{f-1}$. Then, since 14 has no factors $3, 5, 11, 13$ their exponents are at most $1$. Furthermore since $2$ and $7$ divide both divide $14$, I tried finding any combination with different multiplicities, but none yielded $14$ as a Totative.

Was my approach correct or did I miss something?

Rodrigoss
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  • Your conclusion is correct. Here is a simpler argument. – lulu Feb 09 '21 at 23:56
  • And here are some other arguments. – lulu Feb 09 '21 at 23:58
  • Thank you! I tried looking for a post with my question but could not find it. – Rodrigoss Feb 09 '21 at 23:58
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    You shold be wary of the fact that $\varphi(2^a\cdots 13^f)=2^{a+6}3^{b+1}5^c7^d11^{e-1}13^{f-1}$ holds only if $a,b,c,d,e,f>0$, so actually you are just looking for necessary conditions that a $n$ which is divisible by $2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13$ must satisfy in order for its totient function to be $14$ (spoiler: you won't find any). You should investigate the other cases too with the correct formula applied, if you want an actual result, because the conclusions you are drawing only hold in a special case. –  Feb 10 '21 at 00:06
  • It would be straightforward to check each of the six primes $p_i$ in range to see if $\phi(p_i) \mid 14$, and eliminate those which didn't pass this filter. – Joffan Feb 10 '21 at 00:13
  • @Gae.S. would you care to elaborate a little on what other cases should I investigate? And what formula to use? – Rodrigoss Feb 10 '21 at 18:09
  • @Rodrigoss Ah, off the top of my head, I'd say all the $64$ cases that come from imposing some of the six exponents to be $0$ and all the others to be $\ge1$, although the necessity that $7\mid\varphi(n)$ and $\varphi(n)/7=2$ removes a lot of them already. For the formula, you can of course use the fact that, for $a_i\ge1$ and $p_i$ primes, $\varphi\left(\prod_{i=1}^k p_i^{a_i}\right)=\prod_{i=1}^k p^{a_i-1}(p_i-1)$, or equivalently $\varphi(n)=n\prod_{p\mid n}\left(1-\frac1p\right)$. –  Feb 10 '21 at 18:16

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Regarding your method, you limit the primes of interest nicely, but you also need to eliminate from consideration any of those primes which do not pass the basic divisibility test of $\phi(p_i) \mid 14$. This would leave you with only $2$ and $3$ which clearly cannot produce the factor of $7$ needed to make $\phi(n)=14$.

Thinking about this problem separately from your approach, I would start with the key aspects:
a) $14+1 = 15$ is not a prime, and
b) $14=2\cdot 7$

So from (a) we do not arrive at $\phi(n)=14$ directly, and from (b) we need a factor of $7$. However that can only be supplied by a higher power of $7$ - which cannot be present as this would also include a factor of $6$ - or a larger prime (e.g. $29$), which would also include a larger factor not present.

So there is no number $n$ with $\phi(n)=14$.

The OEIS A005277 sequence gives even non-totients; $14$ is the smallest.

Joffan
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