Does knowing $\phi (n)$ help in prime factorization? If you know $\phi (n)$ how can you derive the prime factorization of $n$ from this? For example, $\phi(100) = 40$, but how does $40$ help us come to $2^25^2$?
 A: Let $n = \prod_{i=1}^k p_i^{e_i}$ for distinct primes $p_i$ and nonnegative integers $e_i$.  Then $\varphi(n) = \prod_{i=1}^k \varphi(p_i^{e_i}) = \prod_{i=1}^k p_i^{e_i}\left(1-\frac{1}{p_i}\right)$.  For any $e_i \geq 2$, we find that $p_i | (\varphi(n),n)$.  In your example, $(\varphi(100),100) = (40,100) = 20$ which is divisible by both $2$ and $5$.
If $n$ is squarefree, you get somewhat less information, $(p_i - 1)|\varphi(n)$, so try to factor $\varphi(n)$, add one to each of its divisors (there are generally more of these than factors) and trial divide $n$ by these.
If $n$ is the product of two distinct primes, $p$ and $q$, then $\varphi(n) = (p-1)(q-1) = pq -p -q +1 = n -p -q +1$ which is enough information to find the factorization.  I.e., you know $pq=n$ and $p+q = n+1-\varphi(n)$.  This can be partially extended to three factors, $p$, $q$, and $r$, giving a polynomial whose $pq$, $pr$, and $qr$ terms can be bounded since the roots are all real (i.e., the discriminant is positive).  Not quite as good as a factorization, but may yield a small enough range for sieving to work quickly.
The down side is that if $n$ isn't constructed so as to have a special totient value, the only way to compute $\varphi(n)$ is from $n$'s prime factorization.  Which puts the cart before the horse.
