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The question here is if there exists $x,y,z\in\mathbb Z$ such that$$y^2=x^2-x^3z^2+z-1\label{1}\tag{1}$$other than the trivial solution$$x=0,y^2+1=z\text{ for all }y\in\mathbb Z\label2\tag2$$I know that if there are solutions to$$\left(\dfrac yx\right)^2=z(1-xz)\label{3}\tag{3}$$then those exact same solutions will satisfy $\eqref{1}$ (although I would be missing solutions of the original equation by simplifying, such as $\eqref2$) although plugging $\eqref1$ into Mathematica

Solve[{b^2==a^2-a^3c^2+c-1, -100 <= a <= 100, -100 <= b <= 100, -100 <= c <= 100}, {a, b, c}, Integers]

takes too long to solve compared to the time allowed for free Wolfram Cloud subscriptions and plugging in $\eqref3$ into Mathematica

Solve[{Divide[b^2,a^2]=c(1-ac),-100<=a<=100,-100<=b<=100,-100<=c<=100},{a,b,c},Integers]

will automatically abort. (Probably because it considers the case $x=a=0$, although that's not a problem.)

I am not looking for a Mathematica solution to this. The only reason I used Mathematica in the first place was to try to find solutions that way to see if there are any solutions that it could find that I couldn't by hand.

I am looking for a way that I could find any solutions to the Diophantine equation other than the one trivial solution, or just a proof that there are none other.

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    $\begingroup$ Yes. For example, you can pick any integer $t$, let $x = 1-t^{2}$, $y = t^{3}-t$ and $z = 1$. $\endgroup$
    – Jeremy Rouse
    Commented Feb 27 at 21:35
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    $\begingroup$ $(x, y, z) = (-1, 0, -1), (-1, 0, 0)$. Also $x > 1$ is impossible because $x^3z^2 > x^2 + z - 1$. $\endgroup$
    – Denis Shatrov
    Commented Feb 27 at 21:56
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    $\begingroup$ Why are you calling this a "curve"? It looks like a surface to me. $\endgroup$
    – Joe Silverman
    Commented Feb 28 at 2:38
  • $\begingroup$ @JoeSilverman I think that that was because when graphing the equation, it looked like it was a curve to me, so that's probably why I am calling it a curve. $\endgroup$
    – CrSb0001
    Commented Feb 28 at 2:40

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