## Problem:

Pentagonal numbers are generated by the formula, Pn=n(3n[−]1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 [−] 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk [−] Pj| is minimised; what is the value of D?

5482660

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

public interface EulerSolution{

public String run();

}
/*
* Solution to Project Euler problem 44
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/

public final class p044 implements EulerSolution {

public static void main(String[] args) {
System.out.println(new p044().run());
}

public String run() {
long minD = -1;
// For each upper pentagonal number index, going upward
for (int i = 2; ; i++) {
long pentI = pentagonalNumber(i);
// If the next number down is more than a found difference, then conclude searching
if (minD != -1 && pentI - pentagonalNumber(i - 1) > minD)
break;

// For each lower pentagonal number index, going downward
for (int j = i - 1; j >= 1; j--) {
long pentJ = pentagonalNumber(j);
long diff = pentI - pentJ;
// If the difference is at least as big as a found difference, then stop testing lower pentagonal numbers
if (minD != -1 && diff >= minD)
break;
else if (isPentagonalNumber(pentI + pentJ) && isPentagonalNumber(diff) && (minD == -1 || diff < minD))
minD = diff;  // Found a smaller difference
}
}
return Long.toString(minD);
}

private static long pentagonalNumber(int x) {
if (x <= 0)
throw new IllegalArgumentException();
return (long)x * (x * 3 - 1) >>> 1;
}

private static boolean isPentagonalNumber(long y) {
if (y <= 0)
return false;

/*
* If y = pentagonalNumber(x) = x(3x-1) / 2,
* then by the quadratic formula, the positive solution is x = (sqrt(24y + 1) + 1) / 6.
* There exists a solution for x if and only if both of these conditions hold:
* (24y + 1) is a perfect square, and sqrt(24y + 1) + 1 mod 6 = 0.
* The second condition is equivalent to sqrt(24y + 1) = 5 mod 6.
*/
long temp = y * 24 + 1;
long sqrt = Library.sqrt(temp);
return sqrt * sqrt == temp && sqrt % 6 == 5;
}

}