Below is the syntax highlighted version of Euler.java
from §1.3 Conditionals and Loops.

/****************************************************************************** * Compilation: javac Euler.java * Execution: java Euler n * * Tests whether there are any five positive integers that satisfy * a^5 + b^5 + c^5 + d^5 = e^5. In 1769 Euler conjectured that no such * solutions exists, but his conjecture was disproved in 1966 using * a method like the one below. * * The program reads in an integer command-line argument n and prints * all solutions with a <= b <= c <= d <= e <= n. Restricting attention * to solutions of this form only eliminates duplicates and makes * the program faster since we have many fewer possibilities to try * (675,993,780 vs. 75,937,500,000). * * For further efficiency, we use the break statement to avoid explicitly * enumerating certain tuples (a, b, c, d, e), e.g., if a^5 + b^5 * is already greater than e^5, there is no need to fix specific * values of c and d and compute a^5 + b^5 + c^5 + d^5. On my system, * this decreased the running time from 3 minutes to 35 seconds. * * % java Euler 100 * * % java Euler 150 * 27^5 + 84^5 + 110^5 + 133^5 = 144^5 // takes about 35 seconds * * ******************************************************************************/ public class Euler { public static void main(String[] args) { long n = Long.parseLong(args[0]); long a5, b5, c5, d5, e5; for (long e = 1; e <= n; e++) { e5 = e*e*e*e*e; // restrict search to a <= b <= c <= d <= e for efficiency for (long a = 1; a <= n; a++) { a5 = a*a*a*a*a; if (a5 + a5 + a5 + a5 > e5) break; for (long b = a; b <= n; b++) { b5 = b*b*b*b*b; if (a5 + b5 + b5 + b5 > e5) break; for (long c = b; c <= n; c++) { c5 = c*c*c*c*c; if (a5 + b5 + c5 + c5 > e5) break; for (long d = c; d <= n; d++) { d5 = d*d*d*d*d; if (a5 + b5 + c5 + d5 > e5) break; if (a5 + b5 + c5 + d5 == e5) System.out.println(a + "^5 + " + b + "^5 + " + c + "^5 + " + d + "^5 = " + e + "^5"); } } } } } } }

Last updated: Fri Oct 20 14:12:12 EDT 2017.