# Count number of solutions of x^2 = 1 (mod p) in given range

Given two integers n and p, find the number of integral solutions to x^{2} = 1 (mod p) in the closed interval [1, n].

**Examples:**

Input : n = 10, p = 5 Output : 4 There are four integers that satisfy the equation x^{2}= 1. The numbers are 1, 4, 6 and 9. Input : n = 15, p = 7 Output : 5 There are five integers that satisfy the equation x^{2}= 1. The numbers are 1, 8, 15, 6 and 13.

One simple solution is to go through all numbers from 1 to n. For every number, check if it satisfies the equation. We can avoid going through the whole range. The idea is based on the fact that if a number x satisfies the equation, then all numbers of the form x + i*p also satisfy the equation. We traverse for all numbers from 1 to p and for every number x that satisfies the equation, we find the count of numbers of the form x + i*p. To find the count, we first find the largest number for given x and then add (largest-number – x)/p to the result.

Below is the implementation of the idea.

## C++

`// C++ program to count number of values` `// that satisfy x^2 = 1 mod p where x lies` `// in range [1, n]` `#include<bits/stdc++.h>` `using` `namespace` `std;` `typedef` `long` `long` `ll;` `int` `findCountOfSolutions(` `int` `n, ` `int` `p)` `{` ` ` `// Initialize result` ` ` `ll ans = 0;` ` ` `// Traverse all numbers smaller than` ` ` `// given number p. Note that we don't` ` ` `// traverse from 1 to n, but 1 to p` ` ` `for` `(ll x=1; x<p; x++)` ` ` `{` ` ` `// If x is a solution, then count all` ` ` `// numbers of the form x + i*p such` ` ` `// that x + i*p is in range [1,n]` ` ` `if` `((x*x)%p == 1)` ` ` `{` ` ` `// The largest number in the` ` ` `// form of x + p*i in range` ` ` `// [1, n]` ` ` `ll last = x + p * (n/p);` ` ` `if` `(last > n)` ` ` `last -= p;` ` ` `// Add count of numbers of the form` ` ` `// x + p*i. 1 is added for x itself.` ` ` `ans += ((last-x)/p + 1);` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `int` `main()` `{` ` ` `ll n = 10, p = 5;` ` ` `printf` `(` `"%lld\n"` `, findCountOfSolutions(n, p));` ` ` `return` `0;` `}` |

## Java

`// Java program to count` `// number of values that` `// satisfy x^2 = 1 mod p` `// where x lies in range [1, n]` `import` `java.io.*;` `class` `GFG` `{` `static` `int` `findCountOfSolutions(` `int` `n,` ` ` `int` `p)` `{` ` ` `// Initialize result` ` ` `int` `ans = ` `0` `;` ` ` `// Traverse all numbers` ` ` `// smaller than given` ` ` `// number p. Note that` ` ` `// we don't traverse from` ` ` `// 1 to n, but 1 to p` ` ` `for` `(` `int` `x = ` `1` `; x < p; x++)` ` ` `{` ` ` `// If x is a solution,` ` ` `// then count all numbers` ` ` `// of the form x + i*p` ` ` `// such that x + i*p is` ` ` `// in range [1,n]` ` ` `if` `((x * x) % p == ` `1` `)` ` ` `{` ` ` `// The largest number` ` ` `// in the form of x +` ` ` `// p*i in range [1, n]` ` ` `int` `last = x + p * (n / p);` ` ` `if` `(last > n)` ` ` `last -= p;` ` ` `// Add count of numbers` ` ` `// of the form x + p*i.` ` ` `// 1 is added for x itself.` ` ` `ans += ((last - x) / p + ` `1` `);` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `n = ` `10` `;` ` ` `int` `p = ` `5` `;` ` ` `System.out.println(` ` ` `findCountOfSolutions(n, p));` `}` `}` `// This code is contributed by ajit` |

## Python3

`# Program to count number of` `# values that satisfy x^2 = 1` `# mod p where x lies in range [1, n]` `def` `findCountOfSolutions(n, p):` ` ` ` ` `# Initialize result` ` ` `ans ` `=` `0` `;` ` ` `# Traverse all numbers smaller` ` ` `# than given number p. Note` ` ` `# that we don't traverse from` ` ` `# 1 to n, but 1 to p` ` ` `for` `x ` `in` `range` `(` `1` `, p):` ` ` ` ` `# If x is a solution, then` ` ` `# count all numbers of the` ` ` `# form x + i*p such that` ` ` `# x + i*p is in range [1,n]` ` ` `if` `((x ` `*` `x) ` `%` `p ` `=` `=` `1` `):` ` ` ` ` `# The largest number in the` ` ` `# form of x + p*i in range` ` ` `# [1, n]` ` ` `last ` `=` `x ` `+` `p ` `*` `(n ` `/` `p);` ` ` `if` `(last > n):` ` ` `last ` `-` `=` `p;` ` ` `# Add count of numbers of` ` ` `# the form x + p*i. 1 is` ` ` `# added for x itself.` ` ` `ans ` `+` `=` `((last ` `-` `x) ` `/` `p ` `+` `1` `);` ` ` `return` `int` `(ans);` `# Driver code` `n ` `=` `10` `;` `p ` `=` `5` `;` `print` `(findCountOfSolutions(n, p));` ` ` `# This code is contributed by mits` |

## C#

`// C# program to count` `// number of values that` `// satisfy x^2 = 1 mod p` `// where x lies in range [1, n]` `using` `System;` `class` `GFG` `{` `static` `int` `findCountOfSolutions(` `int` `n,` ` ` `int` `p)` `{` ` ` `// Initialize result` ` ` `int` `ans = 0;` ` ` `// Traverse all numbers` ` ` `// smaller than given` ` ` `// number p. Note that` ` ` `// we don't traverse from` ` ` `// 1 to n, but 1 to p` ` ` `for` `(` `int` `x = 1; x < p; x++)` ` ` `{` ` ` `// If x is a solution,` ` ` `// then count all numbers` ` ` `// of the form x + i*p` ` ` `// such that x + i*p is` ` ` `// in range [1,n]` ` ` `if` `((x * x) % p == 1)` ` ` `{` ` ` `// The largest number` ` ` `// in the form of x +` ` ` `// p*i in range [1, n]` ` ` `int` `last = x + p * (n / p);` ` ` `if` `(last > n)` ` ` `last -= p;` ` ` `// Add count of numbers` ` ` `// of the form x + p*i.` ` ` `// 1 is added for x itself.` ` ` `ans += ((last - x) / p + 1);` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `static` `public` `void` `Main ()` `{` ` ` `int` `n = 10;` ` ` `int` `p = 5;` ` ` `Console.WriteLine(` ` ` `findCountOfSolutions(n, p));` `}` `}` `// This code is contributed by ajit` |

## PHP

`<?php` `// Program to count number of` `// values that satisfy x^2 = 1` `// mod p where x lies in range [1, n]` `function` `findCountOfSolutions(` `$n` `, ` `$p` `)` `{` ` ` `// Initialize result` ` ` `$ans` `= 0;` ` ` `// Traverse all numbers smaller` ` ` `// than given number p. Note` ` ` `// that we don't traverse from` ` ` `// 1 to n, but 1 to p` ` ` `for` `(` `$x` `= 1; ` `$x` `< ` `$p` `; ` `$x` `++)` ` ` `{` ` ` `// If x is a solution, then` ` ` `// count all numbers of the` ` ` `// form x + i*p such that` ` ` `// x + i*p is in range [1,n]` ` ` `if` `((` `$x` `* ` `$x` `) % ` `$p` `== 1)` ` ` `{` ` ` `// The largest number in the` ` ` `// form of x + p*i in range` ` ` `// [1, n]` ` ` `$last` `= ` `$x` `+ ` `$p` `* (` `$n` `/ ` `$p` `);` ` ` `if` `(` `$last` `> ` `$n` `)` ` ` `$last` `-= ` `$p` `;` ` ` `// Add count of numbers of` ` ` `// the form x + p*i. 1 is` ` ` `// added for x itself.` ` ` `$ans` `+= ((` `$last` `- ` `$x` `) / ` `$p` `+ 1);` ` ` `}` ` ` `}` ` ` `return` `$ans` `;` `}` `// Driver code` `$n` `= 10;` `$p` `= 5;` `echo` `findCountOfSolutions(` `$n` `, ` `$p` `);` ` ` `// This code is contributed by ajit` `?>` |

## Javascript

`<script>` `// Javascript program to count number` `// of values that satisfy x^2 = 1 mod p` `// where x lies in range [1, n]` `function` `findCountOfSolutions(n, p)` `{` ` ` ` ` `// Initialize result` ` ` `let ans = 0;` ` ` ` ` `// Traverse all numbers smaller` ` ` `// than given number p. Note that` ` ` `// we don't traverse from 1 to n,` ` ` `// but 1 to p` ` ` `for` `(let x = 1; x < p; x++)` ` ` `{` ` ` ` ` `// If x is a solution,` ` ` `// then count all numbers` ` ` `// of the form x + i*p` ` ` `// such that x + i*p is` ` ` `// in range [1,n]` ` ` `if` `((x * x) % p == 1)` ` ` `{` ` ` ` ` `// The largest number` ` ` `// in the form of x +` ` ` `// p*i in range [1, n]` ` ` `let last = x + p * (n / p);` ` ` ` ` `if` `(last > n)` ` ` `last -= p;` ` ` ` ` `// Add count of numbers` ` ` `// of the form x + p*i.` ` ` `// 1 is added for x itself.` ` ` `ans += ((last - x) / p + 1);` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` ` ` `// Driver code` `let n = 10;` `let p = 5;` `document.write(findCountOfSolutions(n, p));` ` ` `// This code is contributed by susmitakundugoaldanga` ` ` `</script>` |

**Output:**

4

This article is contributed by **Shubham Agrawal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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