Find all perfect square root of n digit

Here given code implementation process.

// C program 
// Find all perfect square root of n digit
#include <stdio.h>
#include <math.h>
int isSquareRoot(int sum)
{
	int value = (int) sqrt(sum);
	if (value *value == sum)
	{
		return 1;
	}
	return 0;
}
// Display Calculated result 
void show(int output[], int n, int sum)
{
	for (int i = 0; i < n; ++i)
	{
		if (i != 0)
		{
			printf("+");
		}
		printf("%d²", output[i]);
	}
	printf(" = %d(%d²)\n ", sum, (int) sqrt(sum));
}
// Find all perfect square root of n digit sum
void perfectSquare(int output[], int n, int counter, int num, int sum)
{
	if (counter == n)
	{
		if (isSquareRoot(sum))
		{
			show(output, n, sum);
		}
		return;
	}
	for (int x = num; x <= 9; ++x)
	{
		output[counter] = x;
		sum += x *x;
		perfectSquare(output, n, counter + 1, x, sum);
		sum -= x *x;
	}
}
// Handles the request of finding digit square sum
void findDigitSquare(int digit)
{
	if (digit <= 0)
	{
		// When digit is less than 1
		return;
	}
	// Assume digit is not exceed the length of maximum numbers
	int output[digit];
	printf("\n Perfect Square of %d digits are \n ", digit);
	perfectSquare(output, digit, 0, 1, 0);
}
int main()
{
	// Test case
	findDigitSquare(3);
	findDigitSquare(4);
	return 0;
}

Output

 Perfect Square of 3 digits are
 1²+2²+2² = 9(3²)
 1²+4²+8² = 81(9²)
 2²+3²+6² = 49(7²)
 2²+4²+4² = 36(6²)
 2²+6²+9² = 121(11²)
 3²+6²+6² = 81(9²)
 4²+4²+7² = 81(9²)
 4²+8²+8² = 144(12²)
 6²+6²+7² = 121(11²)

 Perfect Square of 4 digits are
 1²+1²+1²+1² = 4(2²)
 1²+1²+3²+5² = 36(6²)
 1²+1²+7²+7² = 100(10²)
 1²+2²+2²+4² = 25(5²)
 1²+3²+3²+9² = 100(10²)
 1²+4²+4²+4² = 49(7²)
 1²+5²+5²+7² = 100(10²)
 2²+2²+2²+2² = 16(4²)
 2²+2²+3²+8² = 81(9²)
 2²+2²+4²+5² = 49(7²)
 2²+2²+7²+8² = 121(11²)
 2²+4²+4²+8² = 100(10²)
 2²+4²+5²+6² = 81(9²)
 2²+8²+8²+8² = 196(14²)
 3²+3²+3²+3² = 36(6²)
 3²+5²+9²+9² = 196(14²)
 4²+4²+4²+4² = 64(8²)
 4²+4²+5²+8² = 121(11²)
 4²+5²+8²+8² = 169(13²)
 4²+6²+6²+9² = 169(13²)
 4²+8²+8²+9² = 225(15²)
 5²+5²+5²+5² = 100(10²)
 6²+6²+6²+6² = 144(12²)
 7²+7²+7²+7² = 196(14²)
 8²+8²+8²+8² = 256(16²)
 9²+9²+9²+9² = 324(18²)

/*
  Java Program for
  Find all perfect square root of n digit number
*/

public class DigitSquare
{
   
    public boolean isSquareRoot(int sum)
    {
        int value = (int) Math.sqrt(sum);
        if (value * value == sum)
        {
            return true;
        }
        return false;
    }
    // Display Calculated result 
    public void show(int[] output, int n, int sum)
    {
        for (int i = 0; i < n; ++i)
        {
            if (i != 0)
            {
                System.out.print("+");
            }
            System.out.print(" "+output[i] + "²");
        }
        System.out.print(" = " + sum + "(" + (int) Math.sqrt(sum) + "²)\n");
    }
    // Find all perfect square root of n digit sum
    public void perfectSquare(int[] output, int n, int counter, int num, int sum)
    {
        if (counter == n)
        {
            if (isSquareRoot(sum))
            {
                show(output, n, sum);
            }
            return;
        }
        for (int x = num; x <= 9; ++x)
        {
            output[counter] = x;
            sum += x * x;
            perfectSquare(output, n, counter + 1, x, sum);
            sum -= x * x;
        }
    }
    // Handles the request of finding digit square sum
    public void findDigitSquare(int digit)
    {
        if (digit <= 0)
        {
            // When digit is less than 1
            return;
        }
        // Assume digit is not exceed the length of maximum numbers
        int[] output = new int[digit];
        System.out.print("\n Perfect Square of " + digit + " digits are \n");
        perfectSquare(output, digit, 0, 1, 0);
    }
    public static void main(String[] args)
    {
        DigitSquare task = new DigitSquare();
        // Test case
        task.findDigitSquare(3);
        task.findDigitSquare(4);
    }
}

Output

 Perfect Square of 3 digits are
 1²+ 2²+ 2² = 9(3²)
 1²+ 4²+ 8² = 81(9²)
 2²+ 3²+ 6² = 49(7²)
 2²+ 4²+ 4² = 36(6²)
 2²+ 6²+ 9² = 121(11²)
 3²+ 6²+ 6² = 81(9²)
 4²+ 4²+ 7² = 81(9²)
 4²+ 8²+ 8² = 144(12²)
 6²+ 6²+ 7² = 121(11²)

 Perfect Square of 4 digits are
 1²+ 1²+ 1²+ 1² = 4(2²)
 1²+ 1²+ 3²+ 5² = 36(6²)
 1²+ 1²+ 7²+ 7² = 100(10²)
 1²+ 2²+ 2²+ 4² = 25(5²)
 1²+ 3²+ 3²+ 9² = 100(10²)
 1²+ 4²+ 4²+ 4² = 49(7²)
 1²+ 5²+ 5²+ 7² = 100(10²)
 2²+ 2²+ 2²+ 2² = 16(4²)
 2²+ 2²+ 3²+ 8² = 81(9²)
 2²+ 2²+ 4²+ 5² = 49(7²)
 2²+ 2²+ 7²+ 8² = 121(11²)
 2²+ 4²+ 4²+ 8² = 100(10²)
 2²+ 4²+ 5²+ 6² = 81(9²)
 2²+ 8²+ 8²+ 8² = 196(14²)
 3²+ 3²+ 3²+ 3² = 36(6²)
 3²+ 5²+ 9²+ 9² = 196(14²)
 4²+ 4²+ 4²+ 4² = 64(8²)
 4²+ 4²+ 5²+ 8² = 121(11²)
 4²+ 5²+ 8²+ 8² = 169(13²)
 4²+ 6²+ 6²+ 9² = 169(13²)
 4²+ 8²+ 8²+ 9² = 225(15²)
 5²+ 5²+ 5²+ 5² = 100(10²)
 6²+ 6²+ 6²+ 6² = 144(12²)
 7²+ 7²+ 7²+ 7² = 196(14²)
 8²+ 8²+ 8²+ 8² = 256(16²)
 9²+ 9²+ 9²+ 9² = 324(18²)
// Include header file
#include <iostream>
#include <math.h>
using namespace std;

/*
  C++ Program for
  Find all perfect square root of n digit number
*/

class DigitSquare
{
	public: bool isSquareRoot(int sum)
	{
		int value = (int) sqrt(sum);
		if (value *value == sum)
		{
			return true;
		}
		return false;
	}
	// Display Calculated result
	void show(int output[], int n, int sum)
	{
		for (int i = 0; i < n; ++i)
		{
			if (i != 0)
			{
				cout << "+";
			}
			cout << " " << output[i] << "²";
		}
		cout << " = " << sum << "(" << (int) sqrt(sum) << "²)\n";
	}
	// Find all perfect square root of n digit sum
	void perfectSquare(int output[], int n, int counter, int num, int sum)
	{
		if (counter == n)
		{
			if (this->isSquareRoot(sum))
			{
				this->show(output, n, sum);
			}
			return;
		}
		for (int x = num; x <= 9; ++x)
		{
			output[counter] = x;
			sum += x *x;
			this->perfectSquare(output, n, counter + 1, x, sum);
			sum -= x *x;
		}
	}
	// Handles the request of finding digit square sum
	void findDigitSquare(int digit)
	{
		// When digit is less than 1
		if (digit <= 0)
		{
			return;
		}
		// Assume digit is not exceed the length of maximum numbers
		int output[digit];
		cout << "\n Perfect Square of " << digit << " digits are \n";
		this->perfectSquare(output, digit, 0, 1, 0);
	}
};
int main()
{
	DigitSquare task = DigitSquare();
	// Test case
	task.findDigitSquare(3);
	task.findDigitSquare(4);
	return 0;
}

Output

 Perfect Square of 3 digits are
 1²+ 2²+ 2² = 9(3²)
 1²+ 4²+ 8² = 81(9²)
 2²+ 3²+ 6² = 49(7²)
 2²+ 4²+ 4² = 36(6²)
 2²+ 6²+ 9² = 121(11²)
 3²+ 6²+ 6² = 81(9²)
 4²+ 4²+ 7² = 81(9²)
 4²+ 8²+ 8² = 144(12²)
 6²+ 6²+ 7² = 121(11²)

 Perfect Square of 4 digits are
 1²+ 1²+ 1²+ 1² = 4(2²)
 1²+ 1²+ 3²+ 5² = 36(6²)
 1²+ 1²+ 7²+ 7² = 100(10²)
 1²+ 2²+ 2²+ 4² = 25(5²)
 1²+ 3²+ 3²+ 9² = 100(10²)
 1²+ 4²+ 4²+ 4² = 49(7²)
 1²+ 5²+ 5²+ 7² = 100(10²)
 2²+ 2²+ 2²+ 2² = 16(4²)
 2²+ 2²+ 3²+ 8² = 81(9²)
 2²+ 2²+ 4²+ 5² = 49(7²)
 2²+ 2²+ 7²+ 8² = 121(11²)
 2²+ 4²+ 4²+ 8² = 100(10²)
 2²+ 4²+ 5²+ 6² = 81(9²)
 2²+ 8²+ 8²+ 8² = 196(14²)
 3²+ 3²+ 3²+ 3² = 36(6²)
 3²+ 5²+ 9²+ 9² = 196(14²)
 4²+ 4²+ 4²+ 4² = 64(8²)
 4²+ 4²+ 5²+ 8² = 121(11²)
 4²+ 5²+ 8²+ 8² = 169(13²)
 4²+ 6²+ 6²+ 9² = 169(13²)
 4²+ 8²+ 8²+ 9² = 225(15²)
 5²+ 5²+ 5²+ 5² = 100(10²)
 6²+ 6²+ 6²+ 6² = 144(12²)
 7²+ 7²+ 7²+ 7² = 196(14²)
 8²+ 8²+ 8²+ 8² = 256(16²)
 9²+ 9²+ 9²+ 9² = 324(18²)
// Include namespace system
using System;
/*
  C# Program for
  Find all perfect square root of n digit number
*/
public class DigitSquare
{
	public Boolean isSquareRoot(int sum)
	{
		int value = (int) Math.Sqrt(sum);
		if (value * value == sum)
		{
			return true;
		}
		return false;
	}
	// Display Calculated result
	public void show(int[] output, int n, int sum)
	{
		for (int i = 0; i < n; ++i)
		{
			if (i != 0)
			{
				Console.Write("+");
			}
			Console.Write(" " + output[i] + "²");
		}
		Console.Write(" = " + sum + "(" + (int) Math.Sqrt(sum) + "²)\n");
	}
	// Find all perfect square root of n digit sum
	public void perfectSquare(int[] output, int n, int counter, int num, int sum)
	{
		if (counter == n)
		{
			if (isSquareRoot(sum))
			{
				show(output, n, sum);
			}
			return;
		}
		for (int x = num; x <= 9; ++x)
		{
			output[counter] = x;
			sum += x * x;
			perfectSquare(output, n, counter + 1, x, sum);
			sum -= x * x;
		}
	}
	// Handles the request of finding digit square sum
	public void findDigitSquare(int digit)
	{
		// When digit is less than 1
		if (digit <= 0)
		{
			return;
		}
		// Assume digit is not exceed the length of maximum numbers
		int[] output = new int[digit];
		Console.Write("\n Perfect Square of " + digit + " digits are \n");
		perfectSquare(output, digit, 0, 1, 0);
	}
	public static void Main(String[] args)
	{
		DigitSquare task = new DigitSquare();
		// Test case
		task.findDigitSquare(3);
		task.findDigitSquare(4);
	}
}

Output

 Perfect Square of 3 digits are
 1²+ 2²+ 2² = 9(3²)
 1²+ 4²+ 8² = 81(9²)
 2²+ 3²+ 6² = 49(7²)
 2²+ 4²+ 4² = 36(6²)
 2²+ 6²+ 9² = 121(11²)
 3²+ 6²+ 6² = 81(9²)
 4²+ 4²+ 7² = 81(9²)
 4²+ 8²+ 8² = 144(12²)
 6²+ 6²+ 7² = 121(11²)

 Perfect Square of 4 digits are
 1²+ 1²+ 1²+ 1² = 4(2²)
 1²+ 1²+ 3²+ 5² = 36(6²)
 1²+ 1²+ 7²+ 7² = 100(10²)
 1²+ 2²+ 2²+ 4² = 25(5²)
 1²+ 3²+ 3²+ 9² = 100(10²)
 1²+ 4²+ 4²+ 4² = 49(7²)
 1²+ 5²+ 5²+ 7² = 100(10²)
 2²+ 2²+ 2²+ 2² = 16(4²)
 2²+ 2²+ 3²+ 8² = 81(9²)
 2²+ 2²+ 4²+ 5² = 49(7²)
 2²+ 2²+ 7²+ 8² = 121(11²)
 2²+ 4²+ 4²+ 8² = 100(10²)
 2²+ 4²+ 5²+ 6² = 81(9²)
 2²+ 8²+ 8²+ 8² = 196(14²)
 3²+ 3²+ 3²+ 3² = 36(6²)
 3²+ 5²+ 9²+ 9² = 196(14²)
 4²+ 4²+ 4²+ 4² = 64(8²)
 4²+ 4²+ 5²+ 8² = 121(11²)
 4²+ 5²+ 8²+ 8² = 169(13²)
 4²+ 6²+ 6²+ 9² = 169(13²)
 4²+ 8²+ 8²+ 9² = 225(15²)
 5²+ 5²+ 5²+ 5² = 100(10²)
 6²+ 6²+ 6²+ 6² = 144(12²)
 7²+ 7²+ 7²+ 7² = 196(14²)
 8²+ 8²+ 8²+ 8² = 256(16²)
 9²+ 9²+ 9²+ 9² = 324(18²)
<?php
/*
  Php Program for
  Find all perfect square root of n digit number
*/
class DigitSquare
{
	public	function isSquareRoot($sum)
	{
		$value = (int) sqrt($sum);
		if ($value * $value == $sum)
		{
			return true;
		}
		return false;
	}
	// Display Calculated result
	public	function show($output, $n, $sum)
	{
		for ($i = 0; $i < $n; ++$i)
		{
			if ($i != 0)
			{
				echo "+";
			}
			echo " ". $output[$i] ."²";
		}
		echo " = ". $sum ."(". (int) sqrt($sum) ."²)\n";
	}
	// Find all perfect square root of n digit sum
	public	function perfectSquare($output, $n, $counter, $num, $sum)
	{
		if ($counter == $n)
		{
			if ($this->isSquareRoot($sum))
			{
				$this->show($output, $n, $sum);
			}
			return;
		}
		for ($x = $num; $x <= 9; ++$x)
		{
			$output[$counter] = $x;
			$sum += $x * $x;
			$this->perfectSquare($output, $n, $counter + 1, $x, $sum);
			$sum -= $x * $x;
		}
	}
	// Handles the request of finding digit square sum
	public	function findDigitSquare($digit)
	{
		// When digit is less than 1
		if ($digit <= 0)
		{
			return;
		}
		// Assume digit is not exceed the length of maximum numbers
		$output = array_fill(0, $digit, 0);
		echo "\n Perfect Square of ". $digit ." digits are \n";
		$this->perfectSquare($output, $digit, 0, 1, 0);
	}
}

function main()
{
	$task = new DigitSquare();
	// Test case
	$task->findDigitSquare(3);
	$task->findDigitSquare(4);
}
main();

Output

 Perfect Square of 3 digits are
 1²+ 2²+ 2² = 9(3²)
 1²+ 4²+ 8² = 81(9²)
 2²+ 3²+ 6² = 49(7²)
 2²+ 4²+ 4² = 36(6²)
 2²+ 6²+ 9² = 121(11²)
 3²+ 6²+ 6² = 81(9²)
 4²+ 4²+ 7² = 81(9²)
 4²+ 8²+ 8² = 144(12²)
 6²+ 6²+ 7² = 121(11²)

 Perfect Square of 4 digits are
 1²+ 1²+ 1²+ 1² = 4(2²)
 1²+ 1²+ 3²+ 5² = 36(6²)
 1²+ 1²+ 7²+ 7² = 100(10²)
 1²+ 2²+ 2²+ 4² = 25(5²)
 1²+ 3²+ 3²+ 9² = 100(10²)
 1²+ 4²+ 4²+ 4² = 49(7²)
 1²+ 5²+ 5²+ 7² = 100(10²)
 2²+ 2²+ 2²+ 2² = 16(4²)
 2²+ 2²+ 3²+ 8² = 81(9²)
 2²+ 2²+ 4²+ 5² = 49(7²)
 2²+ 2²+ 7²+ 8² = 121(11²)
 2²+ 4²+ 4²+ 8² = 100(10²)
 2²+ 4²+ 5²+ 6² = 81(9²)
 2²+ 8²+ 8²+ 8² = 196(14²)
 3²+ 3²+ 3²+ 3² = 36(6²)
 3²+ 5²+ 9²+ 9² = 196(14²)
 4²+ 4²+ 4²+ 4² = 64(8²)
 4²+ 4²+ 5²+ 8² = 121(11²)
 4²+ 5²+ 8²+ 8² = 169(13²)
 4²+ 6²+ 6²+ 9² = 169(13²)
 4²+ 8²+ 8²+ 9² = 225(15²)
 5²+ 5²+ 5²+ 5² = 100(10²)
 6²+ 6²+ 6²+ 6² = 144(12²)
 7²+ 7²+ 7²+ 7² = 196(14²)
 8²+ 8²+ 8²+ 8² = 256(16²)
 9²+ 9²+ 9²+ 9² = 324(18²)
/*
  Node Js Program for
  Find all perfect square root of n digit number
*/
class DigitSquare
{
	isSquareRoot(sum)
	{
		var value = parseInt(Math.sqrt(sum));
		if (value * value == sum)
		{
			return true;
		}
		return false;
	}
	// Display Calculated result
	show(output, n, sum)
	{
		for (var i = 0; i < n; ++i)
		{
			if (i != 0)
			{
				process.stdout.write("+");
			}
			process.stdout.write(" " + output[i] + "²");
		}
		process.stdout.write(" = " + sum + "(" + parseInt(Math.sqrt(sum)) + "²)\n");
	}
	// Find all perfect square root of n digit sum
	perfectSquare(output, n, counter, num, sum)
	{
		if (counter == n)
		{
			if (this.isSquareRoot(sum))
			{
				this.show(output, n, sum);
			}
			return;
		}
		for (var x = num; x <= 9; ++x)
		{
			output[counter] = x;
			sum += x * x;
			this.perfectSquare(output, n, counter + 1, x, sum);
			sum -= x * x;
		}
	}
	// Handles the request of finding digit square sum
	findDigitSquare(digit)
	{
		// When digit is less than 1
		if (digit <= 0)
		{
			return;
		}
		// Assume digit is not exceed the length of maximum numbers
		var output = Array(digit).fill(0);
		process.stdout.write("\n Perfect Square of " + digit + " digits are \n");
		this.perfectSquare(output, digit, 0, 1, 0);
	}
}

function main()
{
	var task = new DigitSquare();
	// Test case
	task.findDigitSquare(3);
	task.findDigitSquare(4);
}
main();

Output

 Perfect Square of 3 digits are
 1²+ 2²+ 2² = 9(3²)
 1²+ 4²+ 8² = 81(9²)
 2²+ 3²+ 6² = 49(7²)
 2²+ 4²+ 4² = 36(6²)
 2²+ 6²+ 9² = 121(11²)
 3²+ 6²+ 6² = 81(9²)
 4²+ 4²+ 7² = 81(9²)
 4²+ 8²+ 8² = 144(12²)
 6²+ 6²+ 7² = 121(11²)

 Perfect Square of 4 digits are
 1²+ 1²+ 1²+ 1² = 4(2²)
 1²+ 1²+ 3²+ 5² = 36(6²)
 1²+ 1²+ 7²+ 7² = 100(10²)
 1²+ 2²+ 2²+ 4² = 25(5²)
 1²+ 3²+ 3²+ 9² = 100(10²)
 1²+ 4²+ 4²+ 4² = 49(7²)
 1²+ 5²+ 5²+ 7² = 100(10²)
 2²+ 2²+ 2²+ 2² = 16(4²)
 2²+ 2²+ 3²+ 8² = 81(9²)
 2²+ 2²+ 4²+ 5² = 49(7²)
 2²+ 2²+ 7²+ 8² = 121(11²)
 2²+ 4²+ 4²+ 8² = 100(10²)
 2²+ 4²+ 5²+ 6² = 81(9²)
 2²+ 8²+ 8²+ 8² = 196(14²)
 3²+ 3²+ 3²+ 3² = 36(6²)
 3²+ 5²+ 9²+ 9² = 196(14²)
 4²+ 4²+ 4²+ 4² = 64(8²)
 4²+ 4²+ 5²+ 8² = 121(11²)
 4²+ 5²+ 8²+ 8² = 169(13²)
 4²+ 6²+ 6²+ 9² = 169(13²)
 4²+ 8²+ 8²+ 9² = 225(15²)
 5²+ 5²+ 5²+ 5² = 100(10²)
 6²+ 6²+ 6²+ 6² = 144(12²)
 7²+ 7²+ 7²+ 7² = 196(14²)
 8²+ 8²+ 8²+ 8² = 256(16²)
 9²+ 9²+ 9²+ 9² = 324(18²)
import math
#   Python 3 Program for
#   Find all perfect square root of n digit number

class DigitSquare :
	def isSquareRoot(self, sum) :
		value = int(math.sqrt(sum))
		if (value * value == sum) :
			return True
		
		return False
	
	#  Display Calculated result
	def show(self, output, n, sum) :
		i = 0
		while (i < n) :
			if (i != 0) :
				print("+", end = "")
			
			print("{0}²".format(output[i]),end="")
			i += 1
		
		print(" = {0}({1})² ".format(sum,int(math.sqrt(sum))))
	
	#  Find all perfect square root of n digit sum
	def perfectSquare(self, output, n, counter, num, sum) :
		if (counter == n) :
			if (self.isSquareRoot(sum)) :
				self.show(output, n, sum)
			
			return
		
		x = num
		while (x <= 9) :
			output[counter] = x
			sum += x * x
			self.perfectSquare(output, n, counter + 1, x, sum)
			sum -= x * x
			x += 1
		
	
	#  Handles the request of finding digit square sum
	def findDigitSquare(self, digit) :
		#  When digit is less than 1
		if (digit <= 0) :
			return
		
		#  Assume digit is not exceed the length of maximum numbers
		output = [0] * (digit)
		print("\n Perfect Square of ", digit ," digits are ")
		self.perfectSquare(output, digit, 0, 1, 0)
	

def main() :
	task = DigitSquare()
	#  Test case
	task.findDigitSquare(3)
	task.findDigitSquare(4)

if __name__ == "__main__": main()

Output

 Perfect Square of  3  digits are
1²+2²+2² = 9(3)²
1²+4²+8² = 81(9)²
2²+3²+6² = 49(7)²
2²+4²+4² = 36(6)²
2²+6²+9² = 121(11)²
3²+6²+6² = 81(9)²
4²+4²+7² = 81(9)²
4²+8²+8² = 144(12)²
6²+6²+7² = 121(11)²

 Perfect Square of  4  digits are
1²+1²+1²+1² = 4(2)²
1²+1²+3²+5² = 36(6)²
1²+1²+7²+7² = 100(10)²
1²+2²+2²+4² = 25(5)²
1²+3²+3²+9² = 100(10)²
1²+4²+4²+4² = 49(7)²
1²+5²+5²+7² = 100(10)²
2²+2²+2²+2² = 16(4)²
2²+2²+3²+8² = 81(9)²
2²+2²+4²+5² = 49(7)²
2²+2²+7²+8² = 121(11)²
2²+4²+4²+8² = 100(10)²
2²+4²+5²+6² = 81(9)²
2²+8²+8²+8² = 196(14)²
3²+3²+3²+3² = 36(6)²
3²+5²+9²+9² = 196(14)²
4²+4²+4²+4² = 64(8)²
4²+4²+5²+8² = 121(11)²
4²+5²+8²+8² = 169(13)²
4²+6²+6²+9² = 169(13)²
4²+8²+8²+9² = 225(15)²
5²+5²+5²+5² = 100(10)²
6²+6²+6²+6² = 144(12)²
7²+7²+7²+7² = 196(14)²
8²+8²+8²+8² = 256(16)²
9²+9²+9²+9² = 324(18)²
#   Ruby Program for
#   Find all perfect square root of n digit number

class DigitSquare 
	def isSquareRoot(sum) 
		value = (Math.sqrt(sum)).to_i
		if (value * value == sum) 
			return true
		end

		return false
	end

	#  Display Calculated result
	def show(output, n, sum) 
		i = 0
		while (i < n) 
			if (i != 0) 
				print("+")
			end

			print(" ", output[i] ,"²")
			i += 1
		end

		print(" = ", sum ,"(", (Math.sqrt(sum)).to_i ,"²)\n")
	end

	#  Find all perfect square root of n digit sum
	def perfectSquare(output, n, counter, num, sum) 
		if (counter == n) 
			if (self.isSquareRoot(sum)) 
				self.show(output, n, sum)
			end

			return
		end

		x = num
		while (x <= 9) 
			output[counter] = x
			sum += x * x
			self.perfectSquare(output, n, counter + 1, x, sum)
			sum -= x * x
			x += 1
		end

	end

	#  Handles the request of finding digit square sum
	def findDigitSquare(digit) 
		#  When digit is less than 1
		if (digit <= 0) 
			return
		end

		#  Assume digit is not exceed the length of maximum numbers
		output = Array.new(digit) {0}
		print("\n Perfect Square of ", digit ," digits are \n")
		self.perfectSquare(output, digit, 0, 1, 0)
	end

end

def main() 
	task = DigitSquare.new()
	#  Test case
	task.findDigitSquare(3)
	task.findDigitSquare(4)
end

main()

Output

 Perfect Square of 3 digits are 
 1²+ 2²+ 2² = 9(3²)
 1²+ 4²+ 8² = 81(9²)
 2²+ 3²+ 6² = 49(7²)
 2²+ 4²+ 4² = 36(6²)
 2²+ 6²+ 9² = 121(11²)
 3²+ 6²+ 6² = 81(9²)
 4²+ 4²+ 7² = 81(9²)
 4²+ 8²+ 8² = 144(12²)
 6²+ 6²+ 7² = 121(11²)

 Perfect Square of 4 digits are 
 1²+ 1²+ 1²+ 1² = 4(2²)
 1²+ 1²+ 3²+ 5² = 36(6²)
 1²+ 1²+ 7²+ 7² = 100(10²)
 1²+ 2²+ 2²+ 4² = 25(5²)
 1²+ 3²+ 3²+ 9² = 100(10²)
 1²+ 4²+ 4²+ 4² = 49(7²)
 1²+ 5²+ 5²+ 7² = 100(10²)
 2²+ 2²+ 2²+ 2² = 16(4²)
 2²+ 2²+ 3²+ 8² = 81(9²)
 2²+ 2²+ 4²+ 5² = 49(7²)
 2²+ 2²+ 7²+ 8² = 121(11²)
 2²+ 4²+ 4²+ 8² = 100(10²)
 2²+ 4²+ 5²+ 6² = 81(9²)
 2²+ 8²+ 8²+ 8² = 196(14²)
 3²+ 3²+ 3²+ 3² = 36(6²)
 3²+ 5²+ 9²+ 9² = 196(14²)
 4²+ 4²+ 4²+ 4² = 64(8²)
 4²+ 4²+ 5²+ 8² = 121(11²)
 4²+ 5²+ 8²+ 8² = 169(13²)
 4²+ 6²+ 6²+ 9² = 169(13²)
 4²+ 8²+ 8²+ 9² = 225(15²)
 5²+ 5²+ 5²+ 5² = 100(10²)
 6²+ 6²+ 6²+ 6² = 144(12²)
 7²+ 7²+ 7²+ 7² = 196(14²)
 8²+ 8²+ 8²+ 8² = 256(16²)
 9²+ 9²+ 9²+ 9² = 324(18²)

/*
  Scala Program for
  Find all perfect square root of n digit number
*/
class DigitSquare
{
	def isSquareRoot(sum: Int): Boolean = {
		var value: Int = (Math.sqrt(sum)).toInt;
		if (value * value == sum)
		{
			return true;
		}
		return false;
	}
	// Display Calculated result
	def show(output: Array[Int], n: Int, sum: Int): Unit = {
		var i: Int = 0;
		while (i < n)
		{
			if (i != 0)
			{
				print("+");
			}
			print(" " + output(i) + "²");
			i += 1;
		}
		print(" = " + sum + "(" + (Math.sqrt(sum)).toInt + "²)\n");
	}
	// Find all perfect square root of n digit sum
	def perfectSquare(output: Array[Int], n: Int, counter: Int, num: Int, sum: Int): Unit = {
		if (counter == n)
		{
			if (this.isSquareRoot(sum))
			{
				this.show(output, n, sum);
			}
			return;
		}
		var x: Int = num;
		while (x <= 9)
		{
			output(counter) = x;
			this.perfectSquare(output, n, counter + 1, x, sum + x * x);
			x += 1;
		}
	}
	// Handles the request of finding digit square sum
	def findDigitSquare(digit: Int): Unit = {
		// When digit is less than 1
		if (digit <= 0)
		{
			return;
		}
		// Assume digit is not exceed the length of maximum numbers
		var output: Array[Int] = Array.fill[Int](digit)(0);
		print("\n Perfect Square of " + digit + " digits are \n");
		this.perfectSquare(output, digit, 0, 1, 0);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: DigitSquare = new DigitSquare();
		// Test case
		task.findDigitSquare(3);
		task.findDigitSquare(4);
	}
}

Output

 Perfect Square of 3 digits are
 1²+ 2²+ 2² = 9(3²)
 1²+ 4²+ 8² = 81(9²)
 2²+ 3²+ 6² = 49(7²)
 2²+ 4²+ 4² = 36(6²)
 2²+ 6²+ 9² = 121(11²)
 3²+ 6²+ 6² = 81(9²)
 4²+ 4²+ 7² = 81(9²)
 4²+ 8²+ 8² = 144(12²)
 6²+ 6²+ 7² = 121(11²)

 Perfect Square of 4 digits are
 1²+ 1²+ 1²+ 1² = 4(2²)
 1²+ 1²+ 3²+ 5² = 36(6²)
 1²+ 1²+ 7²+ 7² = 100(10²)
 1²+ 2²+ 2²+ 4² = 25(5²)
 1²+ 3²+ 3²+ 9² = 100(10²)
 1²+ 4²+ 4²+ 4² = 49(7²)
 1²+ 5²+ 5²+ 7² = 100(10²)
 2²+ 2²+ 2²+ 2² = 16(4²)
 2²+ 2²+ 3²+ 8² = 81(9²)
 2²+ 2²+ 4²+ 5² = 49(7²)
 2²+ 2²+ 7²+ 8² = 121(11²)
 2²+ 4²+ 4²+ 8² = 100(10²)
 2²+ 4²+ 5²+ 6² = 81(9²)
 2²+ 8²+ 8²+ 8² = 196(14²)
 3²+ 3²+ 3²+ 3² = 36(6²)
 3²+ 5²+ 9²+ 9² = 196(14²)
 4²+ 4²+ 4²+ 4² = 64(8²)
 4²+ 4²+ 5²+ 8² = 121(11²)
 4²+ 5²+ 8²+ 8² = 169(13²)
 4²+ 6²+ 6²+ 9² = 169(13²)
 4²+ 8²+ 8²+ 9² = 225(15²)
 5²+ 5²+ 5²+ 5² = 100(10²)
 6²+ 6²+ 6²+ 6² = 144(12²)
 7²+ 7²+ 7²+ 7² = 196(14²)
 8²+ 8²+ 8²+ 8² = 256(16²)
 9²+ 9²+ 9²+ 9² = 324(18²)
import Foundation
/*
  Swift 4 Program for
  Find all perfect square root of n digit number
*/
class DigitSquare
{
	func isSquareRoot(_ sum: Int)->Bool
	{
		let value: Int = Int(sqrt(Double(sum)));
	if (value * value == sum)
	{
		return true;
	}
	return false;
}
// Display Calculated result
func show(_ output: [Int], _ n: Int, _ sum: Int)
{
	var i: Int = 0;
	while (i < n)
	{
		if (i  != 0)
		{
			print(terminator: "+");
		}
		print("\(output[i])²",terminator: "");
		i += 1;
	}
	print(" = \(sum) (\(Int(sqrt(Double(sum)))))²");
}
// Find all perfect square root of n digit sum
func perfectSquare(_ output: inout[Int], _ n: Int, _ counter: Int, _ num: Int, _ sum: Int)
{
	if (counter == n)
	{
		if (self.isSquareRoot(sum))
		{
			self.show(output, n, sum);
		}
		return;
	}
	var x: Int = num;
	while (x <= 9)
	{
		output[counter] = x;
		self.perfectSquare(&output, n, counter + 1, x, sum + x * x);
		x += 1;
	}
}
// Handles the request of finding digit square sum
func findDigitSquare(_ digit: Int)
{
	// When digit is less than 1
	if (digit <= 0)
	{
		return;
	}
	// Assume digit is not exceed the length of maximum numbers
	var output: [Int] = Array(repeating: 0, count: digit);
	print("\n Perfect Square of ", digit ," digits are ");
	self.perfectSquare(&output, digit, 0, 1, 0);
}
}
func main()
{
	let task: DigitSquare = DigitSquare();
	// Test case
	task.findDigitSquare(3);
	task.findDigitSquare(4);
}
main();

Output

 Perfect Square of  3  digits are
1²+2²+2² = 9 (3)²
1²+4²+8² = 81 (9)²
2²+3²+6² = 49 (7)²
2²+4²+4² = 36 (6)²
2²+6²+9² = 121 (11)²
3²+6²+6² = 81 (9)²
4²+4²+7² = 81 (9)²
4²+8²+8² = 144 (12)²
6²+6²+7² = 121 (11)²

 Perfect Square of  4  digits are
1²+1²+1²+1² = 4 (2)²
1²+1²+3²+5² = 36 (6)²
1²+1²+7²+7² = 100 (10)²
1²+2²+2²+4² = 25 (5)²
1²+3²+3²+9² = 100 (10)²
1²+4²+4²+4² = 49 (7)²
1²+5²+5²+7² = 100 (10)²
2²+2²+2²+2² = 16 (4)²
2²+2²+3²+8² = 81 (9)²
2²+2²+4²+5² = 49 (7)²
2²+2²+7²+8² = 121 (11)²
2²+4²+4²+8² = 100 (10)²
2²+4²+5²+6² = 81 (9)²
2²+8²+8²+8² = 196 (14)²
3²+3²+3²+3² = 36 (6)²
3²+5²+9²+9² = 196 (14)²
4²+4²+4²+4² = 64 (8)²
4²+4²+5²+8² = 121 (11)²
4²+5²+8²+8² = 169 (13)²
4²+6²+6²+9² = 169 (13)²
4²+8²+8²+9² = 225 (15)²
5²+5²+5²+5² = 100 (10)²
6²+6²+6²+6² = 144 (12)²
7²+7²+7²+7² = 196 (14)²
8²+8²+8²+8² = 256 (16)²
9²+9²+9²+9² = 324 (18)²
/*
  Kotlin Program for
  Find all perfect square root of n digit number
*/
class DigitSquare
{
	fun isSquareRoot(sum: Int): Boolean
	{
		var value: Int = Math.sqrt(sum.toDouble()).toInt();
		if (value * value == sum)
		{
			return true;
		}
		return false;
	}
	// Display Calculated result
	fun show(output: Array <Int> , n: Int, sum: Int): Unit
	{
		var i: Int = 0;
		while (i < n)
		{
			if (i != 0)
			{
				print("+");
			}
			print("" + output[i] + "²");
			i += 1;
		}
		print(" = " + sum + "(" + Math.sqrt(sum.toDouble()).toInt() + "²)\n ");
	}
	// Find all perfect square root of n digit sum
	fun perfectSquare(output: Array < Int > , n: Int, counter: Int, num: Int, sum: Int): Unit
	{
		if (counter == n)
		{
			if (this.isSquareRoot(sum))
			{
				this.show(output, n, sum);
			}
			return;
		}
		var x: Int = num;
		while (x <= 9)
		{
			output[counter] = x;
			this.perfectSquare(output, n, counter + 1, x, sum + x * x);
			x += 1;
		}
	}
	// Handles the request of finding digit square sum
	fun findDigitSquare(digit: Int): Unit
	{
		// When digit is less than 1
		if (digit <= 0)
		{
			return;
		}
		// Assume digit is not exceed the length of maximum numbers
		var output: Array < Int > = Array(digit)
		{
			0
		};
		print("\n Perfect Square of " + digit + " digits are \n ");
		this.perfectSquare(output, digit, 0, 1, 0);
	}
}
fun main(args: Array < String > ): Unit
{
	var task: DigitSquare = DigitSquare();
	// Test case
	task.findDigitSquare(3);
	task.findDigitSquare(4);
}

Output

 Perfect Square of 3 digits are
 1²+2²+2² = 9(3²)
 1²+4²+8² = 81(9²)
 2²+3²+6² = 49(7²)
 2²+4²+4² = 36(6²)
 2²+6²+9² = 121(11²)
 3²+6²+6² = 81(9²)
 4²+4²+7² = 81(9²)
 4²+8²+8² = 144(12²)
 6²+6²+7² = 121(11²)

 Perfect Square of 4 digits are
 1²+1²+1²+1² = 4(2²)
 1²+1²+3²+5² = 36(6²)
 1²+1²+7²+7² = 100(10²)
 1²+2²+2²+4² = 25(5²)
 1²+3²+3²+9² = 100(10²)
 1²+4²+4²+4² = 49(7²)
 1²+5²+5²+7² = 100(10²)
 2²+2²+2²+2² = 16(4²)
 2²+2²+3²+8² = 81(9²)
 2²+2²+4²+5² = 49(7²)
 2²+2²+7²+8² = 121(11²)
 2²+4²+4²+8² = 100(10²)
 2²+4²+5²+6² = 81(9²)
 2²+8²+8²+8² = 196(14²)
 3²+3²+3²+3² = 36(6²)
 3²+5²+9²+9² = 196(14²)
 4²+4²+4²+4² = 64(8²)
 4²+4²+5²+8² = 121(11²)
 4²+5²+8²+8² = 169(13²)
 4²+6²+6²+9² = 169(13²)
 4²+8²+8²+9² = 225(15²)
 5²+5²+5²+5² = 100(10²)
 6²+6²+6²+6² = 144(12²)
 7²+7²+7²+7² = 196(14²)
 8²+8²+8²+8² = 256(16²)
 9²+9²+9²+9² = 324(18²)



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