Proof of Lagrange's TheoremDate: 11/23/2000 at 15:30:00 From: Alex Tse Subject: Lagrange Theorem (Number Theory) I am looking for a proof of Lagrange's Theorem, which states that any positive integer can be expressed as the sum of 4 square numbers. I did find a proof in "Topics from the Theory of Numbers" by Emil Gorosswald. However, the mathematical language used in that book is too advanced for me. Is there a simpler proof? Or is there any persuasive argument which could make me believe that 4 square numbers is the maximum? Thanks very much for your kind attention. Date: 11/25/2000 at 19:17:19 From: Doctor Floor Subject: Re: Lagrange Theorem (Number Theory) Hi, Alex, thanks for writing. The proof of this theorem (where 0 is included with the squares) by Lagrange (1770) can be done as follows: One can check the following identity [1]: (a^2+b^2+c^2+d^2)(A^2+B^2+C^2+D^2) = (aA+bB+cC+dD)^2+(aB-bA+cD-dC)^2+(aC-cA+dB-bD)^2+(aD-dA+bC-cB)^2 so that it is sufficient to prove the theorem for prime numbers. Let p be a prime, which we want to write as sum of four squares. We can assume that p is odd (since 2=1+1+0+0.) First we will show that there is an m such that mp can be written as the sum of four squares. Consider the congruence classes x^2 + 1 (mod p) for x = 0, 1, 2, ..., (p-1)/2 -(y^2) (mod p) for y = 0, 1, 2, ..., (p-1)/2 Each of these two sets contains (p+1)/2 different elements. Together they contain p+1 values, so that there must be x and y such that x^2 + 1 == -y^2 (mod p) and thus x^2 + y^2 + 1 == 0 (mod p) = mp for some m < p. Let's take the minimal m such that a^2 + b^2 + c^2 + d^2 = mp for some a, b, c and d. Now we will show that m = 1 to complete our proof. To do that, suppose that m > 1. Choose A, B, C and D such that -m/2 < A, B, C, D <= m/2 a==A (mod m), b==B (mod m), c==C (mod m) and d==D (mod m) We have A^2 + B^2 + C^2 + D^2 == a^2 + b^2 + c^2 + d^2 == mp == 0 (mod m) A^2 + B^2 + C^2 + D^2 = rm, for some r >= 0 We see that r <= m because r = (A^2+B^2+C^2+D^2)/m <= 4(m/2)^2 / m = m When r = 0, that means that A, B, C and D are all equal to 0. Then a, b, c and d are all multiples of m, so that a^2+b^2+c^2+d^2 is divisible by m^2, and p is divisible by m. This contradicts 1 < m < p. So r > 0. When r = m, that means that A, B, C and D are all equal to m/2. That would for instance mean that a == m/2 (mod m), and thus a = (m/2) + um for some u. From this we see that a^2 = (m/2)^2 + um^2 + u^2*m^2 so that a^2 == (m/2)^2 == m^2/4 (mod m^2). We find the same congruences for b^2, c^2 and d^2. That would again make a^2+b^2+c^2+d^2 divisible by m^2 and lead to a contradiction. So 0 < r < m. Now we get: mp*mr = (a^2+b^2+c^2+d^2)*(A^2+B^2+C^2+D^2) = f^2+g^2+h^2+i^2 for some f, g, h and i by identity [1]. By this identity it is easy to check that f, g, h and i are all congruent to 0 (mod m). But then we have (dividing by m^2) rp = (f/m)^2 + (g/m)^2 + (h/m)^2 + (i/m)^2. This contradicts m being minimal. So our assumption that m > 1 does not hold. And we have proven m = 1 as desired. (I read this proof in a Dutch book, _Getaltheorie voor beginners_, by Frits Beukers.) If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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