How does the prime number algorithm determine if a number is prime or not?

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The prime number algorithm determines if a number is prime by testing divisibility with integers up to the square root of that number. If a number is composite, it can be factored into two integers, both greater than one, meaning at least one of those factors must be less than or equal to the square root. This mathematical principle ensures that checking only up to the square root is sufficient to confirm primality. Additionally, it is noted that only testing the number 2 and the odd integers up to the square root is necessary for efficiency. This method streamlines the process of identifying prime numbers effectively.
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When trying to determine whether a number is prime or not, the following algorithm is often used: Test all numbers up to [sqrt[n]] ([x] is the ceiling of x) to see if any divide evenly into n. If any do, the number is not prime.

My question is, why do you only have to test up to [sqrt[n]]? How does that work mathematically?
 
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If an integer n is composite, then you can factor it n=ab where a and b are both integers greater than 1. We can't have both a>sqrt(n) and b>sqrt(n), otherwise we'd get n>n. If a<=sqrt(n) then any prime dividing a (and we know there's at least one) is a prime factor of n that's less than sqrt(n). If b<=sqrt(n), we likewise get a prime factor of n less than sqrt(n). So if n is composite, it always has a prime factor <=sqrt(n), and it's enough to check these to verify whether n is composite or not.
 
Furthermore, you don't have to test every number up to the sqrt(n), just 2 and all of the odd ones.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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