The method works with an equation of the form

c

_{1}x

^{2}+ c

_{2}xy + c

_{3}y

^{2}= c

_{4}+ c

_{5}x + c

_{6}y

and to keep things easy for me, I'll use

x

^{2}+ 2xy + 3y

^{2}= 4 + 5x + 6y

so I have

c

_{1}= 1, c

_{2}= 2, c

_{3}= 3, c

_{4}= 4, c

_{5}= 5, and c

_{6}= 6

so next I need to calculate

A = (c

_{2}- 2c

_{1})

^{2}+ 4c

_{1}(c

_{2}- c

_{1}- c

_{3}) = -8

B = 2(c

_{2}- 2c

_{1})(c

_{6}- c

_{5}) + 4c

_{5}(c

_{2}- c

_{1}- c

_{3}) = -40

and

C = (c

_{6}- c

_{5})

^{2}- 4c

_{4}(c

_{2}- c

_{1}- c

_{3}) = 33

and I have then the new quadratic Diophantine:

(2A(x+y) - B)

^{2}- 4AS

^{2}= B

^{2}- 4AC

which is

(-16(x+y) + 40)

^{2}+ 32S

^{2}= 2656

and dividing off 16, I have

(-4(x+y) + 10)

^{2}+ 2S

^{2}= 166.

Which has a solution at S=9, giving

-4(x+y) + 10 = +/- 2

and trying the positive first gives x+y = 2, while the negative gives x+y = 3.

Trying the first case, x = 2-y and plugging that into the equation gives

(2-y)

^{2}+ 2(2-y)y + 3y

^{2}= 4 + 5(2-y) + 6y

which is

4 - 4y + y

^{2}+ 4y - 2y

^{2}+ 3y

^{2}= 4 + 10 - 5y + 6y

which is

2y

^{2}-y - 10 = 0,

so y = (1 +/- sqrt(1 + 80))/4 = (1+/-9)/2 = -2 as the other case is a fraction.

Then x=4, so I can try x=4, y=-2, with

x

^{2}+ 2xy + 3y

^{2}= 4 + 5x + 6y

and get

16 + 2(4)(-2) + 3(-2)

^{2}= 4 + 5(4) + 6(-2)

which is 12 = 12, so they balance out as they must. I'll leave the second solution to the reader. Notice there are only two.

That was easy to solve but I'm curious about the next value in the chain, so looking again at

(-4(x+y) + 10)

^{2}+ 2S

^{2}= 166

I now have c

_{1}= 1, c

_{3}=2, and c

_{4}= 166, while all other values are 0, so

A = -8, B = 0, and C = 1992

so the new quadratic Diophantine after some simplifying is:

(-4(-4(x+y)+10 + S))

^{2}+ 2T

^{2}= 3984

where the right hand side is definitely increasing which indicates an infinite chain.

Intriguingly 3984 = 16(3)(83), which is intriguing because -2 must be a quadratic residue modulo each prime or 0 modulo that prime, and notice that 3 - 2 = 1, and 83 - 2 = 81.

The existence condition from the quadratic residues is probably the absolute determinant as to whether or not integer solutions exist, I'd surmise.

James Harris