Sums of Three SquaresDate: 05/18/98 at 00:52:59 From: steve darcy Subject: Sum of three squares Dear Dr Math: What numbers cannot be expressed as the sum of three squares? I will need the theory behind the answer. Many thanks. Date: 05/18/98 at 15:14:56 From: Doctor Rob Subject: Re: Sum of three squares Theory: Integer squares are congruent to 0, 1, or 4 (mod 8). Theorem: Any integer congruent to 7 (mod 8) cannot be represented as the sum of three squares. Proof: Suppose N = 7 (mod 8), and N = x^2 + y^2 + z^2. Then, reducing modulo 8, you must have 7 as a sum of three numbers, each from the set {0,1,4}. Since N is odd, an odd number of the three squares must be odd, and so an odd number of these three numbers must be 1. If all three are 1, the sum is 3, not 7. If one of them is 1, then the other two must be even and add to 6, which is 2 mod 4, while the numbers are 0 mod 4. This is impossible. Corollary: Any integer N of the form 4^m * (8*k + 7) cannot be written as the sum of three squares. Proof: If m > 1, and N = x^2 + y^2 + z^2 = 0 (mod 4), then each of x, y, and z must be even. Then: N/4 = (x/2)^2 + (y/2)^2 + (z/2)^2 is also representable as the sum of three squares. N/4 = 4^(m-1)*(8*k+7) is thus representable as the sum of three squares. Continue the reduction of the exponent of 4 until it is zero, and then you will have concluded that 8*k + 7 is representable, which is false by the theorem. The proof that these are the only integers not so representable was given by Legendre, in _Essai sur la theorie des nombres_ (1798), pp. 202, 398-9, and Gauss, _Disquistiones Arithmeticae_ (1801), Section 291. It depends on the theory of ternary quadratic forms. I do not know the details, but it is very complicated. There is an alternative proof that is more elementary, longer, and also very complicated, which is due to Liouville and Uspensky. This appears in Uspensky and Heaslet, _Elementary Number Theory_ (1939), pp. 465-474. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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