Why do we check up to the square root of a prime number to determine if it is prime?

Question

To test whether a number is prime or not, why do we have to test whether it is divisible only up to the square root of that number?

Answer 1

Let's suppose that the given integer N is not prime,

Then N can be factorized into two factors a and b , 2 <= a, b < N such that N = a*b. Clearly, both of them can't be greater than sqrt(N) simultaneously.

Let us assume without loss of generality that a is smaller.

Now, if you could not find any divisor of N belonging in the range [2, sqrt(N)], what does that mean?

This means that N does not have any divisor in [2, a] as a <= sqrt(N).

Therefore, a = 1 and b = n and hence By definition, N is prime.

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Further reading if you are not satisfied:

Many different combinations of (a, b) may be possible. Let's say they are:

(a₁, b₁), (a₂, b₂), (a₃, b₃), ..... , (a_k, b_k). Without loss of generality, assume a_i < b_i, 1<= i <=k.

Now, to be able to show that N is not prime it is sufficient to show that none of a_i can be factorized further. And we also know that a_i <= sqrt(N) and thus you need to check till sqrt(N) which will cover all a_i. And hence you will be able to conclude whether or not N is prime.

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