Given are distinct odd primes $\displaystyle a,b$. Show that $\displaystyle 2^{ab}-1$ 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.
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 $\displaystyle ab$(under conditions) exist $\displaystyle n$ such that $\displaystyle ab=\phi(n)=p_1\cdot p_2 \cdot p_3 \cdot m$, when $\displaystyle p_1,p_2,p_3$ your primes...
Interesting...