Integer solutions of an equation

Rearranging the equation given, we have $x^2 - (4y+2)x + 6y^2-20y - 29 = 0$ . Solving the quadratic for $x$ ,

$\begin{aligned} x & = \frac {4y+2 \pm \sqrt{(4y+2)^2-4(6y^2-20y-29)}}2 \\ & = \frac {4y+2 \pm \sqrt{120+96y-8y^2}}2 \\ & = 2y+1 \pm \sqrt{30+24y-2y^2} \end{aligned}$

For $x$ to be an integer $30+24y-2y^2 = 2\left(51-(y-6)^2\right)$ must be a perfect square for some integer $y$ . For $30+24y-2y^2$ to be a perfect square, $51-(y-6)^2$ must be even and $(y-6)^2$ odd. And there are only 4 possibilities $(y-6)^2 = 1, 9, 25, 49$ ; $(y-6)^2 > 49$ will make $x$ unreal. Only when $(y-6)^2 = 1, 49$ make $30+24y-2y^2$ a perfect square and we have:

$\begin{cases} (y-6)^2 = 1 \implies \begin{cases} y = 5 & \implies \begin{cases} x = 1 \\ x = 21 \end{cases} \\ y = 7 & \implies \begin{cases} x = 5 \\ x = 25 \end{cases} \end{cases} \\ (y-6)^2 = 49 \implies \begin{cases} y = -1 & \implies \begin{cases} x = -3 \\ x = 1 \end{cases} \\ y = 13 & \implies \begin{cases} x = 25 \\ x = 29 \end{cases} \end{cases} \end{cases}$

There are a total of $\boxed 8$ integer solutions.

$x^2-4xy+6y^2-2x-20y=29$

$x^2-4xy+{\color{#D61F06}4y^2+2y^2}-2x-{\color{#D61F06}24y+4y}=29$

${\color{#D61F06}(x-2y)^2}-2x+4y+2y^2-24y=29$

$(x-2y)^2-{\color{#D61F06}2(x-2y)}{\color{#3D99F6}+1}+2y^2-24y{\color{#3D99F6}+72}=29{\color{#3D99F6}+1+72}$

${\color{#D61F06}((x-2y)-1)^2}+{\color{#D61F06}2(y^2-12y+36)}={\color{#D61F06}102}$

$(x-2y-1)^2+2{\color{#D61F06}(y-6)^2}=102$

It's possible to notice that $(x-2y-1)^2$ and $2(y-6)^2$ are always going to be a positive number, and since the latter is always going to be pair, $x-2y-1$ also has to be pair, and it can't be bigger than 10, else $(x-2y-1)^2$ will be higher than 102, so the possible values of $x-2y-1$ is 0, 2, 4, 6, 8, 10, -2, -4, -6, -8 and -10. If $x-2y-1$ is 0, 4, 6, 8, -4, -6 or -8, then $\frac{102-(x-2y-1)^2}{2}$ won't be a perfect square, so $(y-6)^2$ won't have an integer solution, but if $x-2y-1$ is 2 or -2, $y-6$ will be 7 or -7 and if $x-2y-1$ is 10 or -10, $y-6$ will be 1 or -1. The possible integer values of $y-6$ is 7, -7, 1 or -1, and each of them have two possible integer values of x, so there are $4*2=8$ integer solutions for the equation.

To preserve parity, x must be odd. Let x = 2r - 1. Substituting, 2r^2 -4r + 3y^2 -4yr -9y = 13, so y must be odd. Let y = 2t - 1. Substituting, r^2 + 6t^2 - 4rt - 14t = 1, so r is odd. Let r = 2a - 1. Substituting, 2a^2 -4ta + 3t^2 = 2a + 5t. Writing as a quadratic equation in a, 2a^2 - (4t + 2)a + (3t^2 - 5t) = 0, with solution: 4a = 4t + 2 +/- sqrt(4*(14t -2t^2 + 1). For real solutions, t < 8. We find the following pairs to be valid: (t,a) = (0,0),(0,1),(3,1),(3,6),(4,2),(4,7),(7,7),(7,8). Substituting for x and y, we have: (x,y) =(- 3,-1),(1,-1),(1,5),(21,5),(5,7),(25,7),(25,13),(29,13). Ed Gray