Lagrange's TheoremDate: 02/27/2001 at 06:01:25 From: Andrew Chapman Subject: Lagrange's theorem Hi, I see in your archives you show the proofs of Lagrange's theorem that every positive integer can be expressed as the sum of four squares - but is there an algorithm or some useful method for identifying *which* four squares? (I'm trying to write a computer program, you see...) Thanks for your help! Andrew Date: 02/27/2001 at 16:07:54 From: Doctor Rob Subject: Re: Lagrange's theorem Thanks for writing to Ask Dr. Math, Andrew. Indeed there is. In fact, there are at least three that I know. Suppose you want to write N as a sum of four squares. If N = 0 (mod 4), then write N/4 = a^2 + b^2 + c^2 + d^2, and then N = (2*a)^2 + (2*b)^2 + (2*c)^2 + (2*d)^2. If N = 2 (mod 4), then write N/2 = a^2 + b^2 + c^2 + d^2, and then N = (a+b)^2 + (a-b)^2 + (c+d)^2 + (c-d)^2. If N = 3 (mod 4), then let a = 2*[sqrt(N/4)] - 1, b = 2*[sqrt((N-a^2)/4)] - 1, M = N - a^2 - b^2. You have reduced the problem to one of writing M = 1 (mod 4) as the sum of two squares (see below). If this turns out to be impossible, subtract 2 from b, recompute M, and try again, until you are able to do this. If N = 1 (mod 4), then let a = 2*[sqrt(N/4)], b = 2*[sqrt((N-a^2)/4)], M = N - a^2 - b^2. Again you have reduced the problem to one of writing M = 1 (mod 4) as the sum of two squares (see below). If this turns out to be impossible, subtract 2 from b, recompute M, and try again, until you are able to do this. There are a couple of ways to write M = 1 (mod 4) as the sum of two squares. One is to factor M. If every prime factor of M that is congruent to 3 (mod 4) appears to an even power, this is possible. If any appears to an odd power, this is impossible. Then you can write M = u^2*P, where P is squarefree and has all its prime factors congruent to 1 (mod 4). If you can write P as the sum of two squares, you are done. If you can write each prime factor of P as the sum of two squares, then you can use the identity (w^2 + x^2)*(y^2 + z^2) = (w*y+x*z)^2 + (w*z-x*y)^2 repeatedly to write P as the sum of two squares. That reduces the problem to that of writing a prime number congruent to 1 (mod 4) as the sum of two squares. To solve p = a^2 + b^2, where p is prime, p = 1 (mod 4), first solve x^2 = -1 (mod p). Then start the Euclidean Algorithm with inputs p and x. a and b will be the first two remainders less than sqrt(p). That reduces the problem to solving x^2 = -1 (mod p). You can do this by finding by trial and error a number y such that the Legendre symbol (y/p) = -1. Then x = y^((p-1)/2) (mod p) will be such a solution. Since half the numbers y from 1 to p-1 will have this property, it is quick to find one. A second method is to compose two lists, one of x^2 (mod N), and the other of -1-x^2 (mod N). Make the lists long enough so that there has to be a common element. That will be so if they cover the range 0 < x <= N/2 + 1. Sort both lists, and find a common element. Those values of x[1] and x[2] will give you x[1]^2 + x[2]^2 + 1 = 0 (mod N), m*N = x[1]^2 + x[2]^2 + 1^2 + 0^2, for some positive integer m. Thus m*N has been expressed as a sum of four squares. The trick is to replace m with smaller and smaller multiples until you get m = 1. If m is even, then exactly one of x[1] and x[2] is even. Suppose x[1] is even. Then define y[1] = x[1]/2, y[2] = x[1]/2, y[3] = (x[2]+1)/2, y[4] = (x[2]-1)/2. It is easy to see that now (m/2)*N = y[1]^2 + y[2]^2 + y[3]^2 + y[4]^2, and we have reduced m to a smaller number. If m is odd, and m >= 3, define y[i] = x[i] (mod m), -(m-1)/2 <= y[i] <= (m-1)/2. Then 0 = x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 (mod m), = y[1]^2 + y[2]^2 + y[3]^2 + y[4]^2 (mod m), m*m' = y[1]^2 + y[2]^2 + y[3]^2 + y[4]^2 for some integer m', and m*m' <= 4*(m-1)^2/4 = (m-1)^2, m' <= (m-1)^2/m < m - 1. Also m' > 0. Now (m*m')*(m*N) = (y[1]^2 + y[2]^2 + y[3]^2 + y[4]^2)* (x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2) = z[1]^2 + z[2]^2 + z[3]^2 + z[4]^2, where z[1] = (x[1]*y[1] + x[2]*y[2] + x[3]*y[3] + x[4]*y[4])/m, z[2] = (x[1]*y[2] - x[2]*y[1] + x[3]*y[4] - x[4]*y[3])/m, z[3] = (x[1]*y[3] + x[2]*y[4] - x[3]*y[1] - x[4]*y[2])/m, z[4] = (x[1]*y[4] + x[2]*y[3] - x[3]*y[2] - x[4]*y[1])/m. It can be easily seen that the z[i]'s are integers. Then m'*N = z[1]^2 + z[2]^2 + z[3]^2 + z[4]^2, and m' < m, so we have reduced m. Continue this until you reach m = 1, and you're done. There is another method that avoids factoring M, and also avoids constructing and sorting the two long lists, if either is too difficult. If you need to know it, write again. By the way, I believe the fact that every positive integer can be written as the sum of four squares is called Bachet's Theorem, since he first stated it explicitly in 1621. Lagrange supplied the first proof about a century and a half later. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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