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Find all number $n\in \mathbb N$ such that $\varphi (n)=14$

$\varphi (n)=p_1^{\alpha_1-1}\cdot p_2^{\alpha_2-1}\cdots p_n^{\alpha_n-1}(p_1-1)\cdots (p_n-1)=14$ so number that divide 14 $x|14$ is $x\in \{1,2,7,14\}$ so $p_i-1=1$ or $p_i-1=2$ or $p_i-1=7$ or $p_i-1=14$ $i\in \{1,2,\ldots n \}$ so $p_i=2, p_i=3, p_i=8, p_i=15$ since $8,15$ is not prime number then $p\in \{2,3\}$.

If we say that $n=2^{\alpha}$ then $\varphi (n)=2^{\alpha-1}=14$ so there is not such $\alpha \in \mathbb N$ that $\varphi(n)=14$ so $n\not=2^{\alpha}$

The same is for $n=3^{\beta}$ and $ n=2^{\alpha}\cdot 3^{\beta}$. So there is not exist $n\in \mathbb N$ such that $\varphi(n)=14$

Is this ok?

Marko Škorić
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2 Answers2

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No, it is not ok, since $$\varphi (n)=(p_1-1)(p_2-1)...(p_k-1)p_1^{\alpha_1-1}\cdot p_2^{\alpha_2-1}\cdots p_k^{\alpha_k-1}=14$$ if $$n= p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}$$ Also, in your formula $n$ apears twice in $\varphi (n)$ and at $p_n$.

So $$p_i-1\mid 14 \implies p_i-1\in\{1,2,7,14\}$$

so $p_i\in\{2,3\}$. Thus $n$ is a product of powers of $2$ and $3$. Now we have $$n=2^a3^b\implies 2^a3^{b-1}=14$$ which is impossibile.

nonuser
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Suppose $n=p_1^{x}p_2^{y}p_3^{z}...$,where $p_1,p_2,p_3..$ are primes.Then $\phi(n)=p_1^{x-1}p_2^{y-1}p_3^{z-1}...(p_1-1)(p_2-1)(p_3-1)..$We know that $p_i-1$ can be unity or an even number, as $p_i$ are primes.This means that there can be atmost $2$ primes in the prime factorization of $n$,as otherwise this would imply $4|14$,which is absurd.
CASE 1:$n$ has $1$ prime in its prime factorization
This would mean that $n$ is a prime power, or $p^{x-1}(p-1)=14$.If $(p-1)$ is unity,$p=2$,contradiction.If $(p-1)$ is even this would mean that $3|14$,absurd.

CASE 2:$n$ has 2 primes in its factorization.
This would mean that $14=p^{x-1}q^{y-1}(p-1)(q-1)$ and thus $(p,q)=(2,3)$ or$(p,q)=(3,2)$ A quick check gives no solutions.
Thus we have no solutions.

Venkaat Balaje
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