Apologies if I've done something stupid here, it's around 30 years since I last did a "proof": We want to prove real solution/s for: a*x^2 + b*x*y + c*y^2 = n If solvable then for some real constant y we have: a*x^2 + b*x*y + c*y^2 - n = 0 So there are solutions if: b^2*y^2 - 4*a*(c*y^2 - n) > 0 (b^2 - 4*a*c)*y^2 - 4*a* n > 0 (b^2 - 4*a*c)*y^2 > 4*a*n Because we know (b^2-4*a*c)>0 and y is real then there is a value of y that will satisfy this therefore there is also a value for x to go with this as a solution to: a*x^2 + b*x*y + c*y^2 = n i.e. there is always a solution. On 4 Jun 2014, at 05:41, Neil Sloane wrote:
Does anyone have a program for doing the following:
Given an indefinite binary quadratic form ax^2+bxy+cy^2 (with discriminant b^2-4ac>0 and not a square) and a positive or negative number n, decide if ax^2+bxy+cy^2 = n has a solution.
(If n = p is a prime, one can use the PARI command qfbsolve(Qfb(a,b,c),p) )
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