P451: Added explanation to Java and Python solutions.

Author: nayuki

java/p451.java

@@ -18,6 +18,41 @@ public static void main(String[] args) {
 	}
 
 
+	/* 
+	 * Let n be an arbitrary integer such that n >= 3.
+	 * When we say that the modular inverse of m modulo n equals m itself,
+	 * the formula is m^-1 = m mod n, which is equivalent to m^2 = 1 mod n.
+	 * 
+	 * We know that if n is prime, then m^2 = 1 mod n has exactly two solutions:
+	 * m = 1, n-1. It is easy to verify that these two numbers are solutions.
+	 * The equation factorizes as (m - 1)(m + 1) = 0 mod n. Because n is prime,
+	 * the numbers form a field, and there are no zero divisors (two arbitrary
+	 * non-zero numbers x and y such that xy = 0). Hence 1 and -1 mod n are
+	 * the only possible solutions to the equation. (Note that for the excluded
+	 * special prime case where n = 2, the solutions 1 and -1 are the same number.)
+	 * 
+	 * Suppose we can find the smallest prime factor of n quickly. (Note that if n is
+	 * prime, then the smallest prime factor is n itself.) This can be achieved by
+	 * building a table ahead of time, using a modification of the sieve of Eratosthenes.
+	 * 
+	 * Suppose that for every n' < n, we know the set of solutions to m^2 = 1 mod n'.
+	 * This means whenever we solve the equation for the number n, we save its solutions
+	 * in an ever-growing list, so that when we work on the next value of n we can access
+	 * all possible smaller solutions. This is essentially an argument by strong induction.
+	 * 
+	 * Let p be the smallest prime factor of n. If p = n, then the set of
+	 * solutions is {1, n - 1}, and we are finished with this value of n.
+	 * 
+	 * Otherwise p < n, and obviously n is an integer multiple of p. Because we are looking
+	 * for values of m such that m^2 = 1 mod n, these candidate m values also must satisfy
+	 * m^2 = 1 mod k for any k that divides n (i.e. k is a factor of n). We look at the set
+	 * of solutions for the modulus k = n/p, which has already been solved because k < n.
+	 * We know that any solution modulo n must be congruent to these solutions modulo k.
+	 * Hence we can try to extend and check these old solutions by brute force. Namely, suppose
+	 * m' is a solution modulo k. Then we check the sequence m = m' + 0k, m' + 1k, m' + 2k, ...,
+	 * m' + (p-1)k modulo n. Because p is usually a small number, this isn't a lot of work to do.
+	 */
+	
 	private static final int LIMIT = 20000000;
 
 	private int[] smallestPrimeFactor;

python/p451.py

@@ -9,6 +9,38 @@
 import array, eulerlib, itertools
 
 
+# Let n be an arbitrary integer such that n >= 3.
+# When we say that the modular inverse of m modulo n equals m itself,
+# the formula is m^-1 = m mod n, which is equivalent to m^2 = 1 mod n.
+# 
+# We know that if n is prime, then m^2 = 1 mod n has exactly two solutions:
+# m = 1, n-1. It is easy to verify that these two numbers are solutions.
+# The equation factorizes as (m - 1)(m + 1) = 0 mod n. Because n is prime,
+# the numbers form a field, and there are no zero divisors (two arbitrary
+# non-zero numbers x and y such that xy = 0). Hence 1 and -1 mod n are
+# the only possible solutions to the equation. (Note that for the excluded
+# special prime case where n = 2, the solutions 1 and -1 are the same number.)
+# 
+# Suppose we can find the smallest prime factor of n quickly. (Note that if n is
+# prime, then the smallest prime factor is n itself.) This can be achieved by
+# building a table ahead of time, using a modification of the sieve of Eratosthenes.
+# 
+# Suppose that for every n' < n, we know the set of solutions to m^2 = 1 mod n'.
+# This means whenever we solve the equation for the number n, we save its solutions
+# in an ever-growing list, so that when we work on the next value of n we can access
+# all possible smaller solutions. This is essentially an argument by strong induction.
+# 
+# Let p be the smallest prime factor of n. If p = n, then the set of
+# solutions is {1, n - 1}, and we are finished with this value of n.
+# 
+# Otherwise p < n, and obviously n is an integer multiple of p. Because we are looking
+# for values of m such that m^2 = 1 mod n, these candidate m values also must satisfy
+# m^2 = 1 mod k for any k that divides n (i.e. k is a factor of n). We look at the set
+# of solutions for the modulus k = n/p, which has already been solved because k < n.
+# We know that any solution modulo n must be congruent to these solutions modulo k.
+# Hence we can try to extend and check these old solutions by brute force. Namely, suppose
+# m' is a solution modulo k. Then we check the sequence m = m' + 0k, m' + 1k, m' + 2k, ...,
+# m' + (p-1)k modulo n. Because p is usually a small number, this isn't a lot of work to do.
 def compute():
 	LIMIT = 20000000