I remeember it was the easiest problem an had least mark. The answer is 3 or 4

the question is first a bit irritating, but it is real easy: show first that a is power of 2, then that a is not divisible by 4, thus a=1 or 2. in the first case n=0, so there is an infinite number of divisors , in the second case show that n is prime, thus the number of divisors is 4.

Peter Scholze if n=a=2 and p=3 then n.a has 3 divisors. Many students last year lost 1 mark because of this.

Can't we use Zsigmondy Theorem for this? If we can,how?

By Zsigmondy we show that a must be 2 and p should be a Mersenne prime number, so n must be prime. But an = 2n, has 4 natural divisors if n is different from 2 and 3 divisors if n=2 and we can control dhe n=2 gives p=3.

oiviluhatef wrote: By Zsigmondy we show that a must be 2 and p should be a Mersenne prime number, so n must be prime. But an = 2n, has 4 natural divisors if n is different from 2 and 3 divisors if n=2 and we can control dhe n=2 gives p=3. Can you write the complete solution,please?