primenumber using Math.sqrt function and while loop

I would like to write program to know given number is primenumber using Math.sqrt function and while loop. How can i achieve it. Please advise
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gudii9Asked:
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phoffric\Commented:
If you can find an integer, A > 1 so that N/A = B, an integer, then N is not prime, since N = A * B.
Could it be that both A and B are greater than sqrt(N)?
No.
sqrt(N) * sqrt(N) = N
If A > sqrt(N) and if B > sqrt(N), then A * B > sqrt(N) * sqrt(N) = N.
That means that one of {A, B} must be < sqrt(N).
That means that if a number N is not prime, then there exists an integer X <= sqrt(N), such that X divides evenly into N.
X = 2
while (X <= sqrt(N))
   Is N/X an integer? If so, then N is not prime. Return false.
   else X = X+1
end loop
Return true because we did not find an integer X <= sqrt(N) that divides evenly into N, so N must be prime.
dpearsonCommented:
Would this do?  (You can convert the for loop to a while if you really wish).

Doug

	public static boolean isPrime(int n) {
		for (int i = 2 ; i <= Math.round(Math.sqrt(n)) ; i++) {
			// Can we divide i into n exactly?
			if (n % i == 0)
				return false ;
		}
		// No factors, then it must be prime
		return true ;
	}

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krakatoaCommented:
Doug - right, and he could also test for divisibility by 2 outside the loop, and then start i at 3, and do i=i+2 each time, just to test for odd number divisibility (wouldn't step over prime multiples, but it would cut down the loop iterations, which might show some speed benefits if the tested number was really large). ?
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