There is Lagrange's four-square theorem, which statea that
Given an integer $N$, we can write $N$ as a sum of four squares $A^2+B^2+C^2+D^2=N$.
How can I find a valid solution with time complexity of $O(1)$ or $O(\sqrt N)$? I only have to find one solution.
Here is my $O(1)$ algorithm:
while(count!=4){
solution = sqrt(N)
print solution
N-= solution*solution
count++
}
But it does not works for all $N$, for example it fails for $N=23$. Is there any other solution?
Try this: Legendre's_three-square_theorem
Assume $4\not\mid N$. If not, divide $N$ by $4^k$ and multiply $2^k$ back in the end. Find an $A$ satisfying $N-A^2\not\equiv0,4,7 \mod 8$. It's always valid to choose $A=\lfloor\sqrt{N}\rfloor$ or $\lfloor\sqrt{N}\rfloor-1$. (Choose the even one if $N\equiv 1,5 \mod 8$ and the odd one otherwise.) Now $N-A^2$ is an integer in $O(\sqrt{N})$ and you can find $B,C,D$ in $O(\sqrt{N})$ easily.
Edit: Since $B,C,D\le cN^{\frac{1}{4}}$ for some small constant $c$, we can enumerate two of them (suppose $B$ and $C$) and check whether the corresponding $D$ is an integer. This works in $O(\sqrt{N})$. There might be some faster solutions.
I am going to provide you with 4 links and I hope it will help you.
this one is an online calculator based on an algorithm from math overflow.
http://www.mathcelebrity.com/foursquare.php?num=+178&pl=Show+Lagrange+Four-Square+Notation
the math overflow algorithm can be found here.
and then there is also this algorithm from cs stackexchange
How fast can we find all Four-Square combinations that sum to N?
and if you want to do a bit of work, then there is this algorithm that can find the sum of 2 squares for a given integer. So you would have to divide your number in two ( not necessarily equal part ) and then add up the results to get a 4 square representation.
