If k=ml, then $n=a^{k}-b^{k}=n_{1},n_{2}, n_{1}=(a^{m}-b^{m}),n_{2}=(a^{(l-1)m}+a^{(l-2)m}b^{m}+...+b^{(l-1)m})$. It proved a), because $(n_{1},n_{2})=1$. From a) we have that $2^{qt-1}|\phi(n)$.

sorry , but I can't understand your solution, could u explain it more? [/hide][/quote]

It's not hard to solve this problem after using Zsigmondy's Theorem. Let $c$ have the prime divisors $p_1, p_2, \cdots, p_q.$ Then, consider the numbers $a^{p_i^j} - b^{p_i^j}$ as $i, j$ vary in $[q] \times [t].$ Note that they are increasing as $p_i^j$ increases. Call these numbers tasty. By Zsigmondy's Theorem, each of the tasty numbers has a prime divisor which does not divide any smaller tasty numbers, because $p_i^j \neq 6$ and this statement is trivial when $p_i^j = 2.$ Now, taking these primes gives us $qt$ prime numbers that divide $a^{c^t} - b^{c^t}.$ $\square$ As for $b$, any number with at least $qt$ prime divisors is divisible by $2^{qt - 1}$ because at least $qt-1$ of them are odd, so the problem easily follows. $\square$