Diophantine equation $x^2-xy-y^2=a^2+ab-b^2$ Prove that for any positive integers a and b equation $x^2-xy-y^2=a^2+ab-b^2$ has infinite many solutions in positive integers
My work
I found a solution $(a+b,b)$, but i don't how to prove infinitness of solutions
 A: Similar to "Vieta Jumping" for binary quadratic form $x^2 - k xy + y^2,$ there is an automorphism. In the case of $x^2 - xy - y^2,$ the mapping, which can be repeated any number of times, is
$$ (x,y) \mapsto ( 2x+y, x+y)   $$ 
Note that if the current, $x,y > 0,$ then $2x+y, x+y > 0$ as well. 
The generating matrix for the automorphism group can be found visually in Conway's topograph. Traditional: given $Ax^2 + B xy + C y^2,$ with discriminant $D = B^2 - 4AC$ positive but not a square, every automorphism (with determinant $+1$) comes from a solution to $\tau^2 - D \sigma^2 = 4,$  then matrix
$$
M =
\left(
\begin{array}{cc}
\frac{\tau - B \sigma}{2} & - C \sigma \\
A \sigma & \frac{\tau + B \sigma}{2}
\end{array}
\right)
$$ 
With $A=1, B=-1, C=-1, D=5, \tau = 3, \sigma = 1$ I got
$$
M =
\left(
\begin{array}{cc}
2 & 1 \\
1 & 1
\end{array}
\right)
$$
I had a drawing of this. As we can see the identity matrix as a pair of column vectors (green), we also see the matrix $M$ in the next pair of representations of $1$ ands $-1,$ using the same backwards slant.

A: Above equation shown below:
$(x^2-xy-y^2)=(a^2+ab-b^2)$  --------$(1)$
Equation $(1)$ has parametric solution hence 
it has infinite many numerical solution's.
$(x,y,a,b)=[(8k^2),(10k-5),(8k^2-8k),(6k-5)]$
For, $k=3$ we get:
$(x,y,a,b)=[(72,25,48,13)$
A: Solution of equation.
$$X^2-XY-Y^2=a^2+ab-b^2$$
Can be written using solutions to the Pell equation....
$$p^2-5s^2=\pm1$$
Knowing the first solutions for.... 
$+1--(p;s)--(9;4)$ 
$-1--(p;s)--(2;1)$
The following can be found by the formula.
$$p_2=9p+20s$$
$$s_2=4p+9s$$
Will make a replacement for the simplified entry....
$$k=p^2\pm6ps+5s^2$$
$$t=p^2\pm2ps+5s^2$$
And then the solutions of the equation can be written in the form...
$$X=(a+b)k^2+2bkt+at^2$$
$$Y=bk^2+2akt+(b-a)t^2$$
So the Pell equation has infinitely many solutions, it means for any number $a,b$ - solutions will always be and infinitely many.
