Find all integer solutions for $2(x^2+y^2)+x+y=5xy$

Question

Find all integer solutions for $2(x^2+y^2)+x+y=5xy$

I have been attempting to solve this question for a long period of time but have never achieved anything. I tried to go back to WolframAlpha and it gave me that the integer solutions were $x=y=2, x=y=0$. I tried to make to factor it and have a number left on the RHS so I can find out its factors but could not factor it. I also tried making it into the form of: $(x-a_1)^2+(y-a_2)^2+(\text{ })(\text{ })$ but was unable to determine what would be located inside the empty brackets and the values of $a_1,a_2$. I also tried to multiply the equation by $2$ to get $4x^2=(2x)^2$ Another attempt was assuming that WLOG $x\ge y \iff x=y+a$ which would give me that $a=0$ and from there I could get that $x=y$ and therefore $x=y=0,2$ from basic math. I hope I could get help on this question and thank you anyways.

Answer 1

$$ (2x-y-1)(x-2y+1) = -1 $$ With integers, you must have either $1 \cdot (-1)$ or $(-1) \cdot 1$ Both cases give an intersection of a pair of (non-parallel) lines, so just two points total

Answer 2

Hint

$$2x^2+x(1-5y)+2y^2+y=0$$ is a quadratic equation in $x$

The discriminant $d=(1-5y)^2-8(2y^2+y)=9y^2-18y+1=(3y-3)^2-8$

As $d$ needs to be perfect square, $(3y-3)^2-8=z^2$(say)

$(3y-3-z)(3y-3+z)=8$

As the multiplicands have the same parity, both must be even

$\dfrac{3y-3-z}2\cdot\dfrac{3y-3+z}2=2$

Answer 3

You have a quadratic equation on $x$ with parameter $y$: $$2x^2+x(1-5y)+2y^2+y=0$$ so the discriminat must be prefect square $$ d^2 = (1-5y)^2-8(2y^2+y) = 9y^2-18y+1$$

So we have $$8 = 9(y-1)^2 -d^2 = (3y-3-d)(3y-3+d)$$

and now should be easy...