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I have the following polynomial:

$$P(x,y,z):=9y^2z^2-30x^2z+90xyz+54yz-270x+81\in\mathbb Q[x].$$

It came up in a larger proof, and I would need in order to complete the proof to prove the following result:

Does there exist $(x,y,z,r)\in\mathbb Q^4$ such that $x\ne 0$ and

$$P(x,y,z)=r^2.$$

We can reformulate the problem in the following way:

Does the algebraic variety defined by

$$9Y^2Z^2-30X^2Z+90XYZ+54YZ-270X+81-T^2$$

have a rational point with $X\ne 0$?

I have no idea how to tackle this problem, I have looked up several articles, but nothing seems to apply to this particular question.

Any hints or references would be greatly appreciated.

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  • $\begingroup$ @TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience. $\endgroup$ – E. Joseph Nov 29 '18 at 10:54
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Two obvious solutions are $P(0,0,0)=(\pm9)^2$.

To find more solutions, plugging in $z=0$ yields $$P(x,y,0)=-270x+81,$$ which is a square for $x=\frac{81-t^2}{270}$ for any choice of $t\in\Bbb{Q}$, and any choice of $y\in\Bbb{Q}$.

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