Check for Amicable Pair

Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number. (A proper divisor of a number is a positive factor of that number other than the number itself. 
Examples:

Input : x = 220, y = 284
Output : Yes 
Proper divisors of 220 are 1, 2, 4, 5,
10, 11, 20, 22, 44, 55 and 110. Sum of 
these is 284. Proper divisors of 284 
are 1, 2, 4, 71 and 142 with sum 220.

Input : 1 2
Output :No

The logic is very simple. We compare sum of the proper divisors of both numbers and compare sum for one number with other number.

C++

// CPP program to check if two numbers are
// Amicable or not.
#include <bits/stdc++.h>
using namespace std;

// Function to calculate sum of all
// proper divisors of a given number
int divSum(int n)
{
    // Sum of divisors
    int result = 0;

    // find all divisors which divides 'num'
    for (int i = 2; i <= sqrt(n); i++)
    {
        // if 'i' is divisor of 'n'
        if (n % i == 0)
        {
            // if both divisors are same
            // then add it once else add
            // both
            if (i == (n / i))
                result += i;
            else
                result += (i + n/i);
        }
    }

    // Add 1 and n to result as above loop
    // considers proper divisors greater
    // than 1.
    return (result + 1);
}

// Returns true if x and y are Amicable
// else false.
bool areAmicable(int x, int y)
{
    if (divSum(x) != y)
       return false;

    return (divSum(y) == x);
}

int main() {
    int x = 220, y = 284;
    if (areAmicable(x, y))
       cout << "Yes";
    else
       cout << "No";
    return 0;
}

Java

// JAVA program to check if two numbers are
// Amicable or not.
import java.io.*;

class GFG
{
    // Function to calculate sum of all
    // proper divisors of a given number
    static int divSum(int n)
    {
        // Sum of divisors
        int result = 0;

        // find all divisors which divides 'num'
        for (int i = 2; i <= Math.sqrt(n); i++)
        {
            // if 'i' is divisor of 'n'
            if (n % i == 0)
            {
                // if both divisors are same
                // then add it once else add
                // both
                if (i == (n / i))
                    result += i;
                else
                    result += (i + n / i);
            }
        }

        // Add 1 and n to result as above loop
        // considers proper divisors greater
        // than 1.
        return (result + 1);
    }

    // Returns true if x and y are Amicable
    // else false.
    static boolean areAmicable(int x, int y)
    {
        if (divSum(x) != y)
        return false;

        return (divSum(y) == x);
    }

    public static void main (String[] args)
    {
        int x = 220, y = 284;
        if (areAmicable(x, y))
        System.out.println( "Yes");
        else
        System.out.println("No");

    }
}

// This code is contributed by vt_m.

Python3

# Python program to check
# if two numbers are
# Amicable or not.
import math

# def to calculate sum
# of all proper divisors
# of a given number
def divSum(n) :

    # Sum of divisors
    result = 0

    # find all divisors
    # which divides 'num'
    for i in range(2, int(math.sqrt(n)) + 1) :

        # if 'i' is
        # divisor of 'n'
        if (n % i == 0) :

            # if both divisors are same
            # then add it once else add
            # both
            if (i == int(n / i)) :
                result = result + i
            else :
                result = result +
                         (i + int(n / i))

    # Add 1 and n to result
    # as above loop considers
    # proper divisors greater
    # than 1.
    return (result + 1)

# Returns true if x and y
# are Amicable else false.
def areAmicable(x, y) :

    if (divSum(x) != y) :
        return False

    return (divSum(y) == x)

# Driver Code
x = 220
y = 284
if (areAmicable(x, y)) :
    print ("Yes")
else :
    print ("No")

# This code is contributed by
# Manish Shaw(manishshaw1)

C#

// C# program to check if two numbers are
// Amicable or not.
using System;

class GFG
{
    // Function to calculate sum of all
    // proper divisors of a given number
    static int divSum(int n)
    {
        // Sum of divisors
        int result = 0;

        // find all divisors which divides 'num'
        for (int i = 2; i <= Math.Sqrt(n); i++)
        {
            // if 'i' is divisor of 'n'
            if (n % i == 0)
            {
                // if both divisors are same
                // then add it once else add
                // both
                if (i == (n / i))
                    result += i;
                else
                    result += (i + n / i);
            }
        }

        // Add 1 and n to result as above loop
        // considers proper divisors greater
        // than 1.
        return (result + 1);
    }

    // Returns true if x and y are Amicable
    // else false.
    static bool areAmicable(int x, int y)
    {
        if (divSum(x) != y)
        return false;

        return (divSum(y) == x);
    }

    public static void Main ()
    {
        int x = 220, y = 284;

        if (areAmicable(x, y))
            Console.WriteLine( "Yes");
        else
            Console.WriteLine("No");

    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to check if two
// numbers are Amicable or not.

// Function to calculate sum of all
// proper divisors of a given number
function divSum( $n)
{

    // Sum of divisors
    $result = 0;

    // find all divisors
    // which divides 'num'
    for ($i = 2; $i <= sqrt($n); $i++)
    {

        // if 'i' is divisor of 'n'
        if ($n % $i == 0)
        {

            // if both divisors are same
            // then add it once else add
            // both
            if ($i == ($n / $i))
                $result += $i;
            else
                $result += ($i + $n / $i);
        }
    }

    // Add 1 and n to result
    // as above loop considers
    // proper divisors greater
    // than 1.
    return ($result + 1);
}

// Returns true if x and y
// are Amicable else false.
function areAmicable($x, $y)
{
    if (divSum($x) != $y)
        return false;

    return (divSum($y) == $x);
}

    // Driver Code
    $x = 220;
    $y = 284;
    if (areAmicable($x, $y))
        echo "Yes";
    else
        echo "No";

// This code is contributed by anuj_67.
?>

Javascript

<script>
    // Javascript program to check if two
// numbers are Amicable or not.

// Function to calculate sum of all
// proper divisors of a given number
function divSum(n)
{

    // Sum of divisors
    let result = 0;

    // find all divisors
    // which divides 'num'
    for (let i = 2; i <= Math.sqrt(n); i++)
    {

        // if 'i' is divisor of 'n'
        if (n % i == 0)
        {

            // if both divisors are same
            // then add it once else add
            // both
            if (i == (n / i))
                result += i;
            else
                result += (i + n / i);
        }
    }

    // Add 1 and n to result
    // as above loop considers
    // proper divisors greater
    // than 1.
    return (result + 1);
}

// Returns true if x and y
// are Amicable else false.
function areAmicable(x, y)
{
    if (divSum(x) != y)
        return false;

    return (divSum(y) == x);
}

    // Driver Code
    let x = 220;
    let y = 284;
    if (areAmicable(x, y))
        document.write("Yes");
    else
        document.write("No");

// This code is contributed by _saurabh_jaiswal.

</script>

Output: 

Yes

Time Complexity: O(√n) 
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above