I have to prove that there doesn't exist $n,m\in\mathbb{N}$ that satisfy $\phi(n)=2*13^{2m-1}$.
My first idea is to think of primes in this situation:
to investigate: $$\phi(3),\phi(13^{2m-1}+1).$$
But I am not sure whether this is the right way.
Any help would be appreciated. Thank you in advance.
First note that $2\cdot 13^{2m-1} + 1$ isn't a prime, as it's divisible by $3$. So this means that if such $n$ exists it can't be prime. Moreover it can't be a power of a prime, as then $\phi(n) = (p-1)p^{r-1}$. But then we must have that $p=13$, which wouldn't be of the wanted form. So we have that $n=sk$ for some coprime $s,k$. From this we have that
$$\phi(n) = \phi(s)\phi(k)$$
But now note that the Euler phi function is even except for $2$ ($1$ is the other exception, but we know that $s,k \not = 1$). As $4 \nmid \phi(n)$ we must have that one of the factors is $2$. WLOG let it be $k$. Then we have that $\phi(n) = \phi(s)$. Repeating the same procedure for $s$ yields that $s$ has $2$ as a factor, but note that $s$ isn't even so we got a contradiction.