1. ## Primes

Given are distinct odd primes $a,b$. Show that $2^{ab}-1$ has at least 3 distinct prime divisors.

No idea how to approach this one, so I would gladly use a hint and try to get at it myself. Thank you in advance.

2. In my opinion you should try the Chinese remainder theorem... (But maybe I wrong here)

Or by using Euler's theorem (Euler's theorem - Wikipedia, the free encyclopedia), prove that for every combination of $ab$(under conditions) exist $n$ such that $ab=\phi(n)=p_1\cdot p_2 \cdot p_3 \cdot m$, when $p_1,p_2,p_3$ your primes...

Interesting...

3. Lowest would be 2^(3*5)-1 = 32767: 7,31,151
But beyond my humble abilities to "show".