Question

Prove that among the terms of the progression 3,7,11,...there are infinitely many prime numbers.

Answer 1

Suppose the contrary and let p be the greatest prime number in the given progression.Consider the numbern$=4p!1$. It is not difficult to see that it is a term of the progression (in fact, the given progres-sion contains all positive integers of the form $4k-1$). Thus n must be composite, since $n>p$. Observe that n is not divisible by any prime of the form $4k-1$(all these are factors in $p!$), hence all the prime factors of n are of the form $4k+1$ . The product of several factors of the form $4k+1$ is again of the form $4k+1$, hence $=4k+1$, for some k.This is a contradiction