sum of four squares of an integer

[vickybat]

^^ Did you check that algorithm i posted given by that guy? I think that's a good solution coz it finds multiple possibilities.

 

[sharang.d]

Is four integers a constraint? 'Cause I've come across some test cases where the numbers are getting represented by more than four integers for ex. 2456.
Or, is there any constraint on the number to be represented?

36^2 + 34^2 + 2^2 + 0^2 = 2456

No one read my links

Java:

Sum of squares
Sum of squares source

 

[Desmond]

^^ Have you added keyboard input? How are you giving input?

I'm getting the following error:

Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
	n cannot be resolved to a variable

	at sum.main(sum.java:32)

Sorry, it takes input as an argument.

 

[icebags]

^^ Did you check that algorithm i posted given by that guy? I think that's a good solution coz it finds multiple possibilities.

now i checked.
the guy takes that assumption, but didn't explain why this logic must work. anyways, thanks for the link.

if u want this way, try this ::

z = input val;

i1 = floor(sqrt(z));                        --> like floor of sqrt(10) is 3
i2 = floor( Sqrt(z - i1^2) );
i3 = floor( Sqrt(z - i1^2 - i2^2) );
i4 =  Sqrt(z - i1^2 - i2^2 - i3^2 ) ;

now, as per that logic, its should be : z = i1^2 + i2^2 + i3^2 + i4^2.

hopefully i didn't make mistake.

36^2 + 34^2 + 2^2 + 0^2 = 2456

No one read my links

big huge program and i am bad at mathematics. (it kinda ruined my HS result). sorry man. .

 

[vickybat]

^^ Will try out ur logic mate.

This works i guess. Algorithm isn't mine and i found it over web.

Check it out guys:

[B]public class lagrange{
	public static void sum_of_squares(int n){
	int t1, t2, t;
   
       for (int i = (int) Math.sqrt(n / 4); i * i <= n; i++) {
       
               t1 = n - i * i;
     
       for (int j = (int) Math.sqrt(t1 / 3); j <= i && j * j <= t1; j++) {
            
               t2 = t1 - j * j;
            
       for (int k = (int) Math.sqrt(t2 / 2); k <= j && k * k <= t2; k++) {
               
                t = (int) Math.sqrt(t2 - k * k);
                
           if (t <= k && t * t == t2 - k * k) {
                    System.out.println("(" + i + "^2) + (" + j + "^2) + ("+ k + "^2) + ("+ t +"^2)");
                }
            }
        }
    }
}
            
        
        public static void main (String [] args){
        	sum_of_squares(29);
        
        }
    }
[/B]

Pass any arguements into the sum_of_squares method. It also gives multiple possibilities. Numbers like 23 won't work coz it requires 5 terms
i.e 4^2 + 2^2+ 1^2+ 1^2+ 1^2

Algorithm for that is different. Interestingly, all counter variables used, yield the answer.