Show that there is no integer n with $\phi(n)$ = 14

Question

I did the following proof and I was wondering if its valid. It feels wrong because I didn't actually test the case when purportedly $n$ is not prime, but please feel free to correct me.

Assume there exists $n$ such that $\phi(n) = 14$. Assume $n$ is prime. Then $\phi(n) = n-1$. Then here, n-1 = 14, so n = 15. We know since Euler's totient function is multiplicative that $\phi(xy)$ = $\phi(x)\phi(y)$, so $\phi(15) = \phi(3)\phi(5)$, but alas $\phi(3)\phi(5) = 2\cdot4 = 8 \ne 14$. If $n$ is not prime a similar argument follows since we know then that $n$ must be composed of prime numbers by the prime factorization theorem.

Answer 1

If a prime $p\mid n$, then $p-1\mid\phi(n)$. If $\phi(n)=14$, then, since the divisors of $14$ are $\{1,2,7,14\}$, $p\in\{2,3\}$. Thus, $n=2^a3^b$ and $\phi(n)$ is of the form $2^j3^k$. However, there is no factor of $7$ in $2^j3^k$.

Answer 2

We answer the question, but in a way that is far too convoluted for the particular numerical example of the post. We prove the following result.

Lemma: If $\varphi(n)=2q$, where $q$ is an odd prime, then $2q+1$ is prime.

An immediate consequence is that we cannot have $\varphi(n)=14$. For then $q=7$, but $2q+1$ is not prime.

We now prove the lemma. Suppose that $\varphi(n)=2q$. We divide the analysis into cases.

Case $1$: Maybe $n=2^km$, where $k\ge 3$ and $m$ is odd. Then by the multiplicativity oof $\varphi$, we have $\varphi(n)=\varphi(2^k)\varphi(m)$. This is impossible, since $\varphi(2^k)=2^{k-1}$, which is divisible by $4$. But $2q$ is not divisible by $4$.

Case $2$: Maybe $n=4m$, where $m$ is odd. Then $\varphi(n)=2\varphi(m)$. Since $2q$ is twice an odd number, it follows that $\varphi(m)$ is odd. This is only possible for an odd number $m$ if $m=1$. But then $\varphi(m)=2\ne 2q$.

Case $3$: Maybe $n=2m$ or $n=m$, where $m$ is an odd number. In either case, $\varphi(n)=\varphi(m)$.

Suppose that $m$ can be factored as a product $st$, where $s$ and $t$ are relatively prime, and neither $s$ nor $t$ is equal to $1$. Then $\varphi(s)$ and $\varphi(t)$ are each even, so $\varphi(m)$ is divisible by $4$. But $2q$ isn't, so we cannot have $\varphi(m)=2q$.

It remains to deal with the case where $m$ is a prime power, perhaps equal to $1$. If $m=1$, then $2q=2$, not twice an odd prime.

Now suppose that $m=p^k$, where $p$ is an odd prime, and $k\ge 1$. Then $\varphi(m)=p^{k-1}(p-1)$. This can be twice an odd prime in $2$ ways: (i) $k=1$ and $p-1$ is twice an odd prime, or (ii) $p=3$ and $k=2$.

In Case (i), $2q=p-1$, so $2q+1$ is prime. In Case (ii), $2q=6$, so again $2q+1$ is prime.

Answer 3

You can prove something more generally:

Theorem: For primes $p > q$, there is a solution to $\varphi(n) = pq \iff pq + 1$ prime

Proof: For $\implies$, suppose $n$ is composite, then $n = p_1^{e_1}\cdots p_n^{e_n}$, and we have $\varphi(n) = \prod p_i^{e_i-1}(p_i-1) = pq$. Then, only $p,q$ can divide $n$ but if $p$ divides $n$ then so does $p-1$, which is a contradiction to the RHS being $pq$ unless $p-1 = q$ (otherwise we will have other primes or higher powers of $q$ show up). The only primes for which this happens is $p= 3, q =2$ in which case $pq+1$ is prime. In every other case, we would have prime factors besides $p$ and $q$ in $pq$, a contradiction. Similarly, $q$ cannot divide $n$. Thus, $n$ cannot be composite. Hence, $n$ is prime but then $\varphi(n) = n-1$ so $n - 1= pq \implies n = pq+1$ which we know is prime.

For the other direction, if $pq+1$ is prime then $\varphi(pq+1) = pq$ so we have a solution to $\varphi(n) = pq$.

Finally, $14 = 7 \cdot 2$ and $14+1$ is not prime.

Answer 4

This does not hold. What kind of similar argument are you then talking about?

You can use the fact that $\phi$ is multiplicative. Assume $n = p_1^{a_1}p_2^{a_2}...p_t^{a_t}$, then $\phi(p_1^{a_1}p_2^{a_2}...p_t^{a_t})$ $\phi(p_1^{a_1})\phi(p_2^{a_2})...\phi(p_t^{a_t}) = 2 \cdot 7 = 1 \cdot 14$

Use this to arrive at a contradiction.