### Friday, September 05, 2008

In the ring of integers, given the quadratic expression

c_1*x^2 + c_2*xy + c_3*y^2 = c_4*z^2 + c_5*zx + c_6*zy

where the c's are constants, for solutions to exist it must be true that

((c_2 - 2c_1)^2 + 4c_1*(c_2 - c_1 - c_3))v^2 + (2(c_2 - 2c_1)*(c_6 - c_5) + 4c_5*(c_2 - c_1 - c_3))v + (c_6 - c_5)^2 - 4c_4*(c_2 - c_1 - c_3) = n^2 mod p

for some n, where p is any prime coprime to z for a given solution, when

v = -(x+y)z^{-1} mod p.

For example with x^2 + y^2 = z^2, I have

c_1 = 1, c_2 = 0, c_3 = 1, c_4 = 1, c_5 = 0, and c_6 = 0

which gives

-4v^2 + 8 = n^2 mod p

for every prime coprime to z, for some n (remember ring is ring of integers) when v = -(x+y)z^{-1} mod p.

Making the substitution for v gives

-4(-(x+y)z^{-1})^2 + 8 = n^2 mod p

so

-4(x+y)^2 + 8z^2 = n^2*z^2 mod p

and since x^2 + y^2 = z^2, I can substitute out z, to get

4(x-y)^2 = n^2*z^2 mod p

so the requirement is met, as of course, there are an infinity of integer solutions to x^2 + y^2 = z^2.

And a square was required here because p can be any prime coprime to a solution for z, so an infinite number of primes must work!

Notice that the result also applies to the general diophantine quadratic in 2 variables by making c_1 = 0 and x=1.

The theorem is proven easily using what I call tautological spaces.