Is there any way to represent any number as sum of 4 squares?

Question

Is there any way to represent any number as sum of 4 squares.

For example 29 can be represented as 5^2+2^2+0^2+0^2

I tried the following code but some numbers giving 5terms for example 23 as 4^2+2^2+1^2+1^2+1^2

the code i tried is :

 x=0;
 while(num!=0){
          x=(int)Math.floor(Math.sqrt(num));
         num=num-(x*x);        
}

Answer 1

Unlike what Bohemian said, I solved 23 in 4 terms as follows:

23 = 3^2 + 3^2 + 2^2 + 1^2

And 29 as follows:

29 = 4^2 + 3^2 + 2^2 + 0^2

My logic would start as:

1. Start with the square root of the number - 1. E.g. SQRT(29) = 5 - 1 = 4; This is now our first term.
2. Take value from point 1), square it and add to it again the squared value from point 1) and see if it's greater than N. If it is, decrement the 2nd sum term by 1 and add it the squared value to value from 1).
3. If the previous squared value terms sums are less than N, find next value term and repeat 2) until you have all 4 terms that adds up to N.

Note: This is for your simple case. For complex case like e.g. 323, this might not work.

323 = 17^2 + 4^2 + 3^2 + 3^2

Bear in mind, as you find the x term, the term's value is less than or equal to the x-1 (previous) term's value.

Answer 2

Here is the algorithm code for you

This will give you all possible combinations ...

int n, t1, t2, t;
        n = 29;//Your number

        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)");
                    }
                }
            }
        }