Given two integers P and Q, the task is to find any two integers whose Greatest Common Divisor(GCD) is P and the difference between their squares is Q. If there doesn’t exist any such integers, then print “-1”.
Examples:
Input: P = 3, Q = 27
Output: 6 3
Explanation:
Consider the two number as 6, 3. Now, the GCD(6, 3) = 3 and 6*6 – 3*3 = 27 which satisfies the condition.Input: P = 1, Q = 100
Output: -1
Approach: The given problem can be solved using based on the following observations:
The given equation can also be written as:
=>
=>
Now for an integral solution of the given equation:
(x+y)(x-y) is always an integer
=> (x+y)(x-y) are divisors of Q
Let (x + y) = p1 and (x + y) = p2
be the two equations where p1 & p2 are the divisors of Q
such that p1 * p2 = Q.
Solving for the above two equation we have:
=>
and
and 
From the above calculations, for x and y to be integral, then the sum of divisors must be even. Since there are 4 possible values for two values of x and y as (+x, +y), (+x, -y), (-x, +y) and (-x, -y).
Therefore the total number of possible solution is given by 4*(count pairs of divisors with even sum).
Now among these pairs, find the pair with GCD as P and print the pair. If no such pair exists, print -1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print a valid pair with// the given criteriaint printValidPair(int P, int Q){ // Iterate over the divisors of Q for (int i = 1; i * i <= Q; i++) { // check if Q is a multiple of i if (Q % i == 0) { // L = (A - B) <- 1st equation // R = (A + B) <- 2nd equation int L = i; int R = Q / i; // Calculate value of A int A = (L + R) / 2; // Calculate value of B int B = (R - L) / 2; // As A and B both are integers // so the parity of L and R // should be the same if (L % 2 != R % 2) { continue; } // Check the first condition if (__gcd(A, B) == P) { cout << A << " " << B; return 0; } } } // If no such A, B exist cout << -1; return 0;}// Driver Codeint main(){ int P = 3, Q = 27; printValidPair(P, Q); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to print a valid pair with// the given criteriastatic int printValidPair(int P, int Q){ // Iterate over the divisors of Q for (int i = 1; i * i <= Q; i++) { // check if Q is a multiple of i if (Q % i == 0) { // L = (A - B) <- 1st equation // R = (A + B) <- 2nd equation int L = i; int R = Q / i; // Calculate value of A int A = (L + R) / 2; // Calculate value of B int B = (R - L) / 2; // As A and B both are integers // so the parity of L and R // should be the same if (L % 2 != R % 2) { continue; } // Check the first condition if (__gcd(A, B) == P) { System.out.print(A+ " " + B); return 0; } } } // If no such A, B exist System.out.print(-1); return 0;}static int __gcd(int a, int b) { return b == 0? a:__gcd(b, a % b); } // Driver Codepublic static void main(String[] args){ int P = 3, Q = 27; printValidPair(P, Q);}}// This code is contributed by 29AjayKumar |
Python3
# python program for the above approachimport math# Function to print a valid pair with# the given criteriadef printValidPair(P, Q): # Iterate over the divisors of Q for i in range(1, int(math.sqrt(Q)) + 1): # check if Q is a multiple of i if (Q % i == 0): # L = (A - B) <- 1st equation # R = (A + B) <- 2nd equation L = i R = Q // i # Calculate value of A A = (L + R) // 2 # Calculate value of B B = (R - L) // 2 # As A and B both are integers # so the parity of L and R # should be the same if (L % 2 != R % 2): continue # Check the first condition if (math.gcd(A, B) == P): print(f"{A} {B}") return 0 # If no such A, B exist print(-1) return 0# Driver Codeif __name__ == "__main__": P = 3 Q = 27 printValidPair(P, Q) # This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;class GFG{ // Function to print a valid pair with // the given criteria static int printValidPair(int P, int Q) { // Iterate over the divisors of Q for (int i = 1; i * i <= Q; i++) { // check if Q is a multiple of i if (Q % i == 0) { // L = (A - B) <- 1st equation // R = (A + B) <- 2nd equation int L = i; int R = Q / i; // Calculate value of A int A = (L + R) / 2; // Calculate value of B int B = (R - L) / 2; // As A and B both are integers // so the parity of L and R // should be the same if (L % 2 != R % 2) { continue; } // Check the first condition if (__gcd(A, B) == P) { Console.Write(A + " " + B); return 0; } } } // If no such A, B exist Console.Write(-1); return 0; } static int __gcd(int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver Code public static void Main() { int P = 3, Q = 27; printValidPair(P, Q); }}// This code is contributed by gfgking |
Javascript
<script> // JavaScript code for the above approach // Recursive function to return gcd of a and b function __gcd(a, b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Function to print a valid pair with // the given criteria function printValidPair(P, Q) { // Iterate over the divisors of Q for (let i = 1; i * i <= Q; i++) { // check if Q is a multiple of i if (Q % i == 0) { // L = (A - B) <- 1st equation // R = (A + B) <- 2nd equation let L = i; let R = Q / i; // Calculate value of A let A = (L + R) / 2; // Calculate value of B let B = (R - L) / 2; // As A and B both are integers // so the parity of L and R // should be the same if (L % 2 != R % 2) { continue; } // Check the first condition if (__gcd(A, B) == P) { document.write(A + " " + B); return 0; } } } // If no such A, B exist document.write(-1); return 0; } // Driver Code let P = 3, Q = 27; printValidPair(P, Q); // This code is contributed by Potta Lokesh </script> |
6 3
Time Complexity: O(sqrt(Q))
Auxiliary Space: O(1)
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