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I thought about asking this type of question after I've seen a question in this site which was the following; $$(x+y)(y+z)(x+z)=13xyz$$ We have shown that this equation has infinitely many solutions with trivial solutions that made us plot $x,y,z=0$ and so forth...

However now I wonder if I can have an approach like this;

$$(x+y)(y+z)(x+z)=txyz$$ $x,y,z,t \in \mathbb{Z} \quad t$ is a constant, no constraints about being positive or negative I can comfortably say that if $x=0$ and $y=0$ there are infinitely many $z$'s If $y,z=0$ there are infinitely many $x$'s so, does that prove $$(x+y)(x+z)(y+z)=txyz$$ for $$x,y,z,t \in \mathbb{Z}$$ has infinitely many solutions?

Jyrki Lahtonen
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Deniz Tuna Yalçın
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    Are you trying to find solutions in $x,y,z$ for a fixed value of $t$, or solutions in $x,y,z,t$ (i.e. allowing $t$ to vary too)? – Clive Newstead Sep 07 '17 at 17:31
  • $t$ is a constant yes. And an integer – Deniz Tuna Yalçın Sep 07 '17 at 17:32
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    suppose $x_0,y_0, z_0$ is a solution. Then $kx_0,ky_0,kz_0$ will also be a solution. So, if one solution exists then infinitely many exist. – Doug M Sep 07 '17 at 17:36
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    $z=0, x=-y$ makes both sides zero. – Mark Bennet Sep 07 '17 at 17:37
  • @DougM there are also many different $x_i,y_i,z_i$'s of course and the same situation applies o them all, but is there any way to show a counter example to this? Or to prove that there aren't any counter-examples at all? – Deniz Tuna Yalçın Sep 07 '17 at 17:39
  • Certainly $z=a, y=2a, x=3a, t=10$ works for any $a$. – Joffan Sep 07 '17 at 17:41
  • A single counter-example wouldn't make sense in this case. The negation of "has infinitely many solutions" is "has no solution OR has finitely many solutions". If you could come up for a condition for (x,y,z) to be a solution and prove that there either 0 or finitely many (x,y,z) that satisfy the condition then you would prove the negation. But the negation is clearly not true, as you can see in the other comments. – FullofDill Sep 07 '17 at 17:52
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    Somebody with Pari should convert this elliptic curve (in homogeneous coordinates) into its Weierstrass normal form. I don't know how many non-isogeneous curves you get, but if the Weierstrass equation is in one of the well studied families, then extensive tables are available from various sources. I don't have the time to recall the steps of doing the conversion (and IIRC it has been automated). Adding the EC-tag to attract the right people here. – Jyrki Lahtonen Sep 07 '17 at 18:09

2 Answers2

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The equation given above is shown below:

$(x+y)(x+z)(z+y)=t(xyz)$

For $t=8$ there is a numerical solution $(x,y,z)=(1,-6,15)$

Also for $t=13$ there is a parametric solution & is given below:

$x=90k^2-120k+40$

$y=27k^2-36k+12$

$z=135k^2-180k+60$

for $k=0$ we get another solution $(x,y,z)=(10,3,15)$

Sam
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  • Hello, how did you find a relation between $k$ and $x,y,z$? – Deniz Tuna Yalçın Sep 14 '17 at 05:31
  • Your solution has a common polynomial factor $$x=90k^2-120k+40 = 10,(3k-2)^2\ y=27k^2-36k+12 = 3,(3k-2)^2\ z=135k^2-180k+60 = 15,(3k-2)^2$$ hence is just scaling. If $x,y,z$ is a solution, then $\beta x, \beta y, \beta z$ is also a solution, as commented by Doug M above. – Tito Piezas III Feb 13 '18 at 11:46
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We have known solution $(x,y,z)=(3,2,10)$ since $(z/y)=5$, take $z=5y$, after substitution & simplifying we get:

$6(x+y)(x+5y)=65xy$

After parametrising above, at $(x,y)=((3+t),(2+kt))$ we get the result.

Sam
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