Parametrization of solutions of diophantine equation

Solution 1:

Equation if we write in the General form:

$$aX^2+bXY+cY^2=eZ^2+jZW+tW^2$$

If in this equation there any equivalent to a quadratic form in which the root is an integer.

$$q=\sqrt{b^2+4a(e+j+t-c)}$$

Then there are solutions. They can be written by making the replacement.

$$x=(b(2(e+j+t)-b)+4ac)s-(b+2a)(j+2t)k$$

$$y=(b^2+4c(e+j+t-a))s^2-2(b+2c)(j+2t)sk+(j^2+4t(a+b+c-e))k^2$$

Then decisions can be recorded and they are as follows:

$$X=(b-2(e+j+t-c)\pm{q})p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$

$$+(((2(e+j+t-c)-b)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$

$$***$$

$$Y=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(2(e+j+t-a)-b)s)\pm{x})pn+$$

$$+(((b+2a)\pm{q})y+2((j+2t)k-(2(e+j+t-a)-b)s)x)n^2$$

$$***$$

$$Z=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$

$$+(((b+2a)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$

$$***$$

$$W=(\pm{q}-(b+2a))p^2+2(q((2(a+b+c-e)-j)k-(b+2c)s)\pm{x})pn+$$

$$+(((b+2a)\pm{q})y+2((2(a+b+c-e)-j)k-(b+2c)s)x)n^2$$

$p,n,k,s $ - integers asked us.