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joro
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I think it have. Probably infinitely many.

\begin{aligned} x =& 6882627592338442563 \\ y =& 4866752642924153522 \\ a =& 4096 \\ b =& -8192 \\ c =& 1 \\ d =& -2 \\ t =& 4 \\ \end{aligned}

Found this way.

Fix $c=1,d=-2$ and solve the Pell equation with large $x,y$.

Then $a=2^{2^t-t}$ and $b= -2 \cdot 2^{2^t-t}$ solves the first equation.

Added

So $x^2 - 2 y^2=1$ have arbitrary large solutions. Fix $t$.

Then $ 2^{2^t-t} (x^2 - 2 y^2) = 2^{2^t-t}$.

I think it have. Probably infinitely many.

\begin{aligned} x =& 6882627592338442563 \\ y =& 4866752642924153522 \\ a =& 4096 \\ b =& -8192 \\ c =& 1 \\ d =& -2 \\ t =& 4 \\ \end{aligned}

Found this way.

Fix $c=1,d=-2$ and solve the Pell equation with large $x,y$.

Then $a=2^{2^t-t}$ and $b= -2 \cdot 2^{2^t-t}$ solves the first equation.

I think it have. Probably infinitely many.

\begin{aligned} x =& 6882627592338442563 \\ y =& 4866752642924153522 \\ a =& 4096 \\ b =& -8192 \\ c =& 1 \\ d =& -2 \\ t =& 4 \\ \end{aligned}

Found this way.

Fix $c=1,d=-2$ and solve the Pell equation with large $x,y$.

Then $a=2^{2^t-t}$ and $b= -2 \cdot 2^{2^t-t}$ solves the first equation.

Added

So $x^2 - 2 y^2=1$ have arbitrary large solutions. Fix $t$.

Then $ 2^{2^t-t} (x^2 - 2 y^2) = 2^{2^t-t}$.

Source Link
joro
  • 25.4k
  • 10
  • 66
  • 121

I think it have. Probably infinitely many.

\begin{aligned} x =& 6882627592338442563 \\ y =& 4866752642924153522 \\ a =& 4096 \\ b =& -8192 \\ c =& 1 \\ d =& -2 \\ t =& 4 \\ \end{aligned}

Found this way.

Fix $c=1,d=-2$ and solve the Pell equation with large $x,y$.

Then $a=2^{2^t-t}$ and $b= -2 \cdot 2^{2^t-t}$ solves the first equation.