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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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## 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 ` `?> ` |

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**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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