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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Answer 2

It's all really just basic uses of Factorization and Square Roots.

It may appear to be abstract, but in reality it simply lies with the fact that a non-prime-number's maximum possible factorial would have to be its square root because:

sqrroot(n) * sqrroot(n) = n.

Given that, if any whole number above 1 and below or up to sqrroot(n) divides evenly into n, then n cannot be a prime number.

Pseudo-code example:

i = 2;

is_prime = true;

while loop (i <= sqrroot(n))
{
  if (n % i == 0)
  {
    is_prime = false;
    exit while;
  }
  ++i;
}

Answer 3

Let's say we have a number "a", which is not prime [not prime/composite number means - a number which can be divided evenly by numbers other than 1 or itself. For example, 6 can be divided evenly by 2, or by 3, as well as by 1 or 6].

6 = 1 × 6 or 6 = 2 × 3

So now if "a" is not prime then it can be divided by two other numbers and let's say those numbers are "b" and "c". Which means

a=b*c.

Now if "b" or "c" , any of them is greater than square root of "a "than multiplication of "b" & "c" will be greater than "a".

So, "b" or "c" is always <= square root of "a" to prove the equation "a=b*c".

Because of the above reason, when we test if a number is prime or not, we only check until square root of that number.

Answer 4

Let's say m = sqrt(n) then m × m = n. Now if n is not a prime then n can be written as n = a × b, so m × m = a × b. Notice that m is a real number whereas n, a and b are natural numbers.

Now there can be 3 cases:

1. a > m ⇒ b < m
2. a = m ⇒ b = m
3. a < m ⇒ b > m

In all 3 cases, min(a, b) ≤ m. Hence if we search till m, we are bound to find at least one factor of n, which is enough to show that n is not prime.

Answer 5

So to check whether a number N is Prime or not. We need to only check if N is divisible by numbers<=SQROOT(N). This is because, if we factor N into any 2 factors say X and Y, ie. N=XY. Each of X and Y cannot be less than SQROOT(N) because then, XY < N Each of X and Y cannot be greater than SQROOT(N) because then, X*Y > N

Therefore one factor must be less than or equal to SQROOT(N) ( while the other factor is greater than or equal to SQROOT(N) ). So to check if N is Prime we need only check those numbers <= SQROOT(N).

Answer 6

Any composite number is a product of primes.

Let say n = p1 * p2, where p2 > p1 and they are primes.

If n % p1 === 0 then n is a composite number.

If n % p2 === 0 then guess what n % p1 === 0 as well!

So there is no way that if n % p2 === 0 but n % p1 !== 0 at the same time. In other words if a composite number n can be divided evenly by p2,p3...pi (its greater factor) it must be divided by its lowest factor p1 too. It turns out that the lowest factor p1 <= Math.square(n) is always true.