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.

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    $\begingroup$ For the second you can note that it is $(2x-y)^2 = 9$. $\endgroup$ – quid Jun 9 '16 at 10:41
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    $\begingroup$ For the first one $y^2=\frac{x+6}{x-2}$, which means $x-2$ is a divisor of $x+6$. $\endgroup$ – Ghartal Jun 9 '16 at 10:41
  • $\begingroup$ i don't see how you have a form of ax²+bx+c=0 sure the -6 is c but but the other terms $\endgroup$ – Michelle_B Jun 9 '16 at 10:42
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    $\begingroup$ 1. When does $(x-2)$ divide $(x+6)$? 2. 9 is a square. $\endgroup$ – Aravind Jun 9 '16 at 10:42

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 $(x,y)=(t,\pm3+2t),\ t\in\mathbb{Z}$.

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  • $\begingroup$ I don't quite get how you go from $2x-y = +- 3$ To the last part immediately $\endgroup$ – Michelle_B Jun 9 '16 at 12:26
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    $\begingroup$ Rearrange it to $y=\pm3+2x$ then parameterize it by letting $x=t$. $\endgroup$ – Ian Miller Jun 9 '16 at 12:40

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