Given an **integer n**, our task is to check whether number n can be represented by the product of two squares. If it is possible then print “yes” otherwise print “no”.**Examples :**

Input:n = 144Output:YesExplanation:

The given number 144 can be represented as 2^{2}* 6^{2}= 144.

Input:n = 25Output:NoExplanation:

The given number 25 cannot be represented as product of two square numbers.

**Naive Approach:**

To solve the problem mentioned above the naive method is to use the **Brute Force method**. Use two for loop iterating till n and each time we will check whether the product of the square of both numbers is equal to N. If we find such a combination then we will print Yes otherwise No.

Below is the implementation of above approach:

## C++

`// C++ implementation to Check whether a number can` `// be represented by the product of two squares` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if there exist two` `// numbers product of whose squares is n.` `bool` `prodSquare(` `int` `n)` `{` ` ` `for` `(` `long` `i = 2; i * i <= n; i++)` ` ` `for` `(` `long` `j = 2; j <= n; j++)` ` ` `// check whether the product of the square` ` ` `// of both numbers is equal to N` ` ` `if` `(i * i * j * j == n)` ` ` `return` `true` `;` ` ` `return` `false` `;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 25;` ` ` `if` `(prodSquare(n))` ` ` `cout << ` `"Yes"` `;` ` ` `else` ` ` `cout << ` `"No"` `;` `}` |

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## Java

`// Java implementation to check whether a number can` `// be represented by the product of two squares` `class` `GFG{` `// Function to check if there exist two` `// numbers product of whose squares is n.` `static` `boolean` `prodSquare(` `int` `n)` `{` ` ` `for` `(` `long` `i = ` `2` `; i * i <= n; i++)` ` ` `for` `(` `long` `j = ` `2` `; j <= n; j++)` ` ` `// Check whether the product of the square` ` ` `// of both numbers is equal to N` ` ` `if` `(i * i * j * j == n)` ` ` `return` `true` `;` ` ` ` ` `return` `false` `;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `25` `;` ` ` `if` `(prodSquare(n))` ` ` `System.out.print(` `"Yes"` `);` ` ` `else` ` ` `System.out.print(` `"No"` `);` `}` `}` `// This code is contributed by gauravrajput1` |

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## Python3

`# Python3 implementation to check whether ` `# a number can be represented by the` `# product of two squares` `# Function to check if there exist two` `# numbers product of whose squares is n.` `def` `prodSquare(n):` ` ` ` ` `for` `i ` `in` `range` `(` `2` `, (n) ` `+` `1` `):` ` ` `if` `(i ` `*` `i < (n ` `+` `1` `)):` ` ` `for` `j ` `in` `range` `(` `2` `, n ` `+` `1` `):` ` ` ` ` `# Check whether the product ` ` ` `# of the square of both` ` ` `# numbers is equal to N` ` ` `if` `((i ` `*` `i ` `*` `j ` `*` `j) ` `=` `=` `n):` ` ` `return` `True` `;` ` ` `return` `False` `;` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `n ` `=` `25` `;` ` ` ` ` `if` `(prodSquare(n)):` ` ` `print` `(` `"Yes"` `);` ` ` `else` `:` ` ` `print` `(` `"No"` `);` `# This code is contributed by Rajput-Ji` |

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## C#

`// C# implementation to check whether ` `// a number can be represented by` `// the product of two squares` `using` `System;` `class` `GFG{` `// Function to check if there` `// exist two numbers product ` `// of whose squares is n.` `static` `bool` `prodSquare(` `int` `n)` `{` ` ` `for` `(` `long` `i = 2; i * i <= n; i++)` ` ` `for` `(` `long` `j = 2; j <= n; j++)` ` ` ` ` `// Check whether the product ` ` ` `// of the square of both ` ` ` `// numbers is equal to N` ` ` `if` `(i * i * j * j == n)` ` ` `return` `true` `;` ` ` ` ` `return` `false` `;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `n = 25;` ` ` ` ` `if` `(prodSquare(n))` ` ` `Console.Write(` `"Yes"` `);` ` ` `else` ` ` `Console.Write(` `"No"` `);` `}` `}` `// This code is contributed by sapnasingh4991` |

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**Output:**

No

**Time Complexity:** O(n)

**Efficient Approach:**

To optimize the above method, use a **hashmap** where we will store the squares of number till sqrt(n) and each time we will search for (n / sqrt(i)) in the hashmap if it exists then return Yes else return No.

## C++

`// C++ implementation to Check whether a number can` `// be represented by the product of two squares` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if there exist two` `// numbers product of whose squares is n` `bool` `prodSquare(` `int` `n)` `{` ` ` `// Initialize map` ` ` `unordered_map<` `float` `, ` `float` `> s;` ` ` `for` `(` `int` `i = 2; i * i <= n; ++i) {` ` ` `// Store square value in hashmap` ` ` `s[i * i] = 1;` ` ` `if` `(s.find(n / (i * i)) != s.end())` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 25;` ` ` `if` `(prodSquare(n))` ` ` `cout << ` `"Yes"` `;` ` ` `else` ` ` `cout << ` `"No"` `;` `}` |

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## Java

`// Java implementation to check whether ` `// a number can be represented by the` `// product of two squares` `import` `java.util.*;` `class` `GFG{` `// Function to check if there exist two` `// numbers product of whose squares is n` `static` `boolean` `prodSquare(` `int` `n)` `{` ` ` ` ` `// Initialize map` ` ` `HashMap<Float, Float> s = ` `new` `HashMap<Float, Float>();` ` ` `for` `(` `int` `i = ` `2` `; i * i <= n; ++i)` ` ` `{` ` ` ` ` `// Store square value in hashmap` ` ` `s.put((` `float` `)(i * i), (` `float` `) ` `1` `);` ` ` ` ` `if` `(s.containsKey((` `float` `) n / (i * i)))` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `25` `;` ` ` `if` `(prodSquare(n))` ` ` `System.out.print(` `"Yes"` `);` ` ` `else` ` ` `System.out.print(` `"No"` `);` `}` `}` `// This code is contributed by 29AjayKumar` |

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## Python3

`# Python3 implementation to check whether` `# a number can be represented by the ` `# product of two squares ` `# Function to check if there exist two ` `# numbers product of whose squares is n ` `def` `prodSquare(n):` ` ` `# Initialize dict/map` ` ` `s ` `=` `dict` `()` ` ` `i ` `=` `2` ` ` `while` `(i ` `*` `i <` `=` `n):` ` ` `# Store square value in hashmap` ` ` `s[i ` `*` `i] ` `=` `1` ` ` `if` `((n ` `/` `/` `(i ` `*` `i)) ` `in` `s):` ` ` `return` `True` ` ` `i ` `+` `=` `1` ` ` `return` `False` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `n ` `=` `25` ` ` `if` `(prodSquare(n)):` ` ` `print` `(` `"Yes"` `)` ` ` `else` `:` ` ` `print` `(` `"No"` `)` `# This code is contributed by himanshu77` |

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## C#

`// C# implementation to check whether ` `// a number can be represented by the` `// product of two squares` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` `// Function to check if there exist two` `// numbers product of whose squares is n` `static` `bool` `prodSquare(` `int` `n)` `{` ` ` ` ` `// Initialize map` ` ` `Dictionary<` `float` `, ` ` ` `float` `> s = ` `new` `Dictionary<` `float` `, ` ` ` `float` `>();` ` ` `for` `(` `int` `i = 2; i * i <= n; ++i)` ` ` `{` ` ` ` ` `// Store square value in hashmap` ` ` `s.Add((` `float` `)(i * i), (` `float` `) 1);` ` ` ` ` `if` `(s.ContainsKey((` `float` `) n / (i * i)))` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `n = 25;` ` ` `if` `(prodSquare(n))` ` ` `Console.Write(` `"Yes"` `);` ` ` `else` ` ` `Console.Write(` `"No"` `);` `}` `}` `// This code is contributed by amal kumar choubey` |

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**Output:**

No

**Time Complexity:** O(sqrt n)

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