Project Euler > Problem 123 > Prime square remainders (Java Solution)

Problem:

Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn[−]1)n + (pn+1)n is divided by pn2.

For example, when n = 3, p3 = 5, and 43 + 63 = 280 [≡] 5 mod 25.

The least value of n for which the remainder first exceeds 109 is 7037.

Find the least value of n for which the remainder first exceeds 1010.

Solution:

4179871

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 23
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p023 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p023().run());
	}
	
	
	private static final int LIMIT = 28123;
	
	private boolean[] isAbundant = new boolean[LIMIT + 1];
	
	public String run() {
		// Compute look-up table
		for (int i = 1; i < isAbundant.length; i++)
			isAbundant[i] = isAbundant(i);
		
		int sum = 0;
		for (int i = 1; i <= LIMIT; i++) {
			if (!isSumOf2Abundants(i))
				sum += i;
		}
		return Integer.toString(sum);
	}
	
	
	private boolean isSumOf2Abundants(int n) {
		for (int i = 0; i <= n; i++) {
			if (isAbundant[i] && isAbundant[n - i])
				return true;
		}
		return false;
	}
	
	
	private static boolean isAbundant(int n) {
		if (n < 1)
			throw new IllegalArgumentException();
		
		int sum = 1;  // Sum of factors less than n
		int end = Library.sqrt(n);
		for (int i = 2; i <= end; i++) {
			if (n % i == 0)
				sum += i + n / i;
		}
		if (end * end == n)
			sum -= end;
		return sum > n;
	}
	
}