A conjecture on numbers coprime to its Euler's totient function

I was investigating the natural numbers $$n$$ such that $$\gcd(n,\phi(n)) = 1$$ where $$\phi(n)$$ is the Euler totient function. Clearly $$\phi(n)$$ is even for $$n > 2$$ hence $$\gcd(n,\phi(n)) \ge 2$$ if $$n$$ is even. Again if $$n = p$$ is an odd prime then $$\phi(p) = p-1$$ which is trivially co-prime to $$p$$. Hence all non-trivial $$n$$ such that $$\gcd(n,\phi(n)) = 1$$ must be odd composites. Apart from $$1$$ and the trivial set of primes, the sequence of composite numbers with this property are $$15, 33, 35,51,65,69,77, 85,87, 91, 95, \ldots$$ I observed the following.

Conjecture: If $$n$$ is an odd composite number such that $$\gcd(n,\phi(n)) = 1$$ then the number of divisors of $$n$$ is a perfect power of $$2$$.

Can this be proved or disproved?

• Emm! As you observed numbers with such property are $15,33,35,51,65,69,77,85,87,91,95$, and so on. We can see that all of these numbers are basically of the form $pq$ for some prime $p$ and $q$ and all these have exactly 4 divisors. Not sure if that helps, but yeah some observation :) – crskhr Sep 22 at 6:09
• @crskhr This is not true in general. $255$ is the smallest number with this property which has more than $8$ divisors. – Nilotpal Kanti Sinha Sep 22 at 6:12
• No! You misunderstood me. $255=pqr$ which again is product of $3$ primes. What i was trying to hint at is the following: If such a $n$ exists, then $n$ is the product of distinct primes, each with exponent 1. – crskhr Sep 22 at 6:14

If $$\gcd(n,\phi(n))=1$$, then $$n$$ must be square-free.
To see this, assume $$p^2|n$$ for some prime $$p$$. Then $$p|\phi(n)$$, and so $$p|\gcd(n,\phi(n))$$.
If $$n=p_1\cdots p_m$$ where $$p_1,\ldots,p_m$$ are distinct primes, then the divisors of $$n$$ are all numbers formed by the product of a subset of these, and there are $$2^m$$ such subset (including the whole set which results in $$n$$ itself, and the empty set which gives 1).