Question

To find whether N is a prime number we only need to look for all numbers less or equal to sqrt(N). Why is that? I am writing a C code so trying to understand a reason behind it.

Answer 1

N is prime if it is a positive integer which is divisible by exactly two positive integers, 1 and N. Since a number’s divisors cannot be larger than that number, this gives rise to a simple primality test:

• If an integer N, greater than 1, is not divisible by any integer in the range [2, N-1], then N is prime. Otherwise, N is not prime.

However, it would be nice to modify this test to make it faster. So let us investigate.

Note that the divisors of N occur in pairs. If N is divisible by a number M, then it is also divisible by N/M. For instance, 12 is divisble by 6, and so also by 2. Furthermore, if M >= sqrt(N), then N/M <= sqrt(N).

This means that if no numbers less than or equal to sqrt(N) divide N, no numbers greater than sqrt(N) divide N either (excepting 1 and N themselves), otherwise a contradiction would arise.

So we have a better test:

• If an integer N, greater than 1, is not divisible by any integer in the range [2, sqrt(N)], then N is prime. Otherwise, N is not prime.

if you consider the reasoning above, you should see that a number which passes this test also passes the first test, and a number which fails this test also fails the first test. The tests are therefore equivalent.