Does knowing the totient of a number help factoring it? Edit: The quoted question addresses only numbers of the form $p^a q^b$, I asked a general question for arbitrary $n$.

If $n$ is a prime or a product of 2 primes then knowing its totient $\varphi(n)$ allows us immediately to find the prime factorization of $n$.
How about a general case? Does knowing $\varphi(n)$


*

*give us a way how to find the prime factorization of $n$,

*help as find a prime factor of $n$, or at least

*help at in finding any factor of $n$? (This turns out to be obvious.)

In fact, it is a well-known observation in cryptography that just knowing a multiple of $\phi(n)$ helps greatly in factoring $n$, regardless of the number of prime factors (if there's only one or two primes dividing $n$ then this has already been covered).
Suppose $m$ is a multiple of $\phi(n)$.  Then if you factor out enough powers of $2$ from $m$, there must exist a divisor $t = m/(2^r)$ such that $\lambda(n) \mid 2t$ but $\lambda(n) \nmid t$.  (Here, $\lambda(n)$ is the Carmichael function, which will necessarily be even, unless $n$ is trivially small.)
It will then be the case that for some bases $b$, $b$ will be a quadratic residue for some prime $p \mid n$ but not for a different $q \mid n$.  In this case, taking the GCD $(b^t-1,n)$ will produce a non-trivial factor of $n$.
One simply has to randomly try different values of $b$ (the expected number of tries is finite), as well as different choices of $t$ (there are at most $\log(n)$ possibilities, and one can use successive squaring to efficiently cover them).
To get all the prime factors of $n$, just keep repeating the process (we still have a multiple of $\phi$ for each of factors found above).
