c

_{1}x

^{2}+ c

_{2}xy + c

_{3}y

^{2}= c

_{4}+ c

_{5}x + c

_{6}y

Here x and y are the two unknowns to be figured out. The base result comes from this post from September 2008. But checking published posts, was referenced talking a general method for reducing binary quadratic Diophantine equations on this blog in this post in May 2011.

And copying from this post, where also note that with my method for reducing can get to the general reduced form:

u

^{2}- Dv

^{2}= C

Where u and v are unknowns. And while I've talked about with C=1 as the two conics equation before, the more general also gives two so that is just the unary case. Letters don't matter of course and like to show as:

x

^{2}- Dy

^{2}= F

where all variables are non-zero integers. And yeah a LOT of abstraction, in that progression, where now you can solve for x and y modularly. Looks like I figured that out in September of 2012.

With a non-zero integer N for which a residue m exists where--m

^{2}= D mod N, and r, any residue modulo N for which Fr

^{-1}mod N exists then a solution is:

2x = r + Fr

^{-1}mod N and 2my = Fr

^{-1}- r mod N

And use that to solve for the modular inverse.

r

r

^{-1}= (n-1)(r + 2my_{0}) - 2md mod NWhere y

_{0}is chosen as is m, with m not equal to r, and n and d are to be determined. They are found from:

**2mdF**

_{0}= [F_{0}(n-1) - 1](r + 2my_{0}) mod Nand

**F**

_{0}= r(r+2my_{0}) mod N---------------------------------------------------

And felt an urge to put all together to kind of see that progression from the most general form, to abstraction to the more basic two conics form, and then to a solution for the modular inverse.

So you end up going from something to do with binary quadratic Diophantine equations to something more general than them.

James Harris