Solve the following diophantic equations

Question

I can't seem to find the solution to two problems in my textbook. They ask us to solve the diophantic equations: 1)

1. $xy²-2y²-x-6=0$

2. $4x²-4xy+y²-9=0$

I tried several things but these two just don't work out in my head. Even a step in the right direction would help.

Answer 1

Rewrite the first as: $x=\frac{2y^2+6}{y^2-1}=\frac{2(y^2-1)+8}{y^2-1}=2+\frac{8}{y^2-1}$.

Hence $y^2-1$ must equal $\pm1$, $\pm2$, $\pm4$ or $\pm8$. The only integer solution to this is $y^2-1=-1$ or $y^2-1=8$ giving $y=0,\pm3$ and hence $(x,y)=(-6,0),(3,-3),(3,3)$.

For the second one you can factorize it as: $(2x-y)^2=9$ and hence $2x-y=\pm3$ and hence $(x,y)=(t,\pm3+2t),\ t\in\mathbb{Z}$.