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Suppose there are $m,n\in \mathbb{N}$ such that $\varphi(n)=2\cdot 7^{5m+4}$.

Then $\varphi(n)=\varphi(p_1^{k_1}p_2^{k_2}...p_r^{k_r})=\varphi(p_1^{k_1})\varphi(p_2^{k_2})...\varphi(p_r^{k_r})=p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)=2\cdot 7^{5m+4}$

In another words, $2=\frac{p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)}{7^{5m+4}}$. This means that one of $p_i$'s is 2 with $k_i=2$. WLOG, assume that $p_1=2$ and $k_1=2$. Then $p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)=7^{5m+4}$ but this is not possible since we must have that $p_i=7$ and $p_i-1=6$ also appears.

Does this make sense?

likeAvirgin
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    That is correct. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Mar 06 '21 at 10:57
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    "one of $p_i$'s is 2 with $k_i=2$". There are other possibilities. For example, if $p_1=3$ and $k_1=1$, then $p_1^{k_1-1}(p_1-1)=2$. – mathlove Mar 06 '21 at 11:13
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    You have not ruled out the possibility of a prime $p$ of the form $2\cdot 7^{5m+4}+1$ which would have the desired totient value. However, this can easily be ruled out since $2\cdot 7^{5m+4}+1$ is divisible by $3$ for all $m$ – Peter Mar 06 '21 at 11:44
  • @mathlove No, then 3 would also appear. But next to one 2, other factors must be all 7's. – likeAvirgin Mar 07 '21 at 22:26
  • You should write that in your proof. You are claiming "This means that one of $p_i$'s is 2 with $k_i=2$" without any proof. – mathlove Mar 08 '21 at 06:13

1 Answers1

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I would suggest the following approach :

If $p$ is a prime factor of $n$, then $p-1$ divides $\varphi(n)=2\cdot 7^{5m+4}$ The only divisors $d$ of $2\cdot 7^{5m+4}$ for which $d+1$ is prime, are $1$ and $2$, since $2\cdot 7^k+1$ is divisible by $3$ and greater than $3$ for every positive integer $k$, whereas $7^k+1$ is obviously not prime for positive integer $k$. Hence $n$ cannot have a prime factor exceeding $3$ and hence its totient value cannot be divisible by $7$.

Peter
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