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For a given $t\geq4$, does the following system of equations have a solution over the integers? $$ax^2+by^2=2^{2^t-t}$$$$cx^2+dy^2=1$$$$0<|ta|^2,|tb|^2,|tc|^2,|td|^2<|x|,|y|$$

If so, how to parametrize the solutions and find them?

For a given $x,y:|x|,|y|<B$, how many such $a,b,c,d$ are there?

Is $gcd(a,b)=1$ possible?

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  • $\begingroup$ Why are you interested in this specific system of diophantine equations? $\endgroup$ – Stefan Kohl Nov 16 '13 at 9:59
  • $\begingroup$ I would be surprised if there is a solution for any given $B$. I tried some attempts in Mathematica and the equations did not work out even at $t=3$ with no limit on the value of $B$. $\endgroup$ – T.... Nov 16 '13 at 10:03
  • $\begingroup$ If $\gcd(c,d) \ne 1$ the second equation doesn't have solution... $\endgroup$ – joro Nov 16 '13 at 16:36
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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}$.

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  • $\begingroup$ does the same trick work for $t=3$? (Asking since mathematica came back empty handed). $\endgroup$ – T.... Nov 16 '13 at 16:00
  • $\begingroup$ does the minimum number of bits in $x$ and $y$ always grow double exponential($2^{2^t}$) or faster in $t$? Could there be a smaller solution? $\endgroup$ – T.... Nov 16 '13 at 16:04
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    $\begingroup$ @JAS I edited the answer. There might be smaller solution. Basically $x,y$ do not depend on $t$ at all, for fixed t they can be arbitrary large. $\endgroup$ – joro Nov 16 '13 at 16:11
  • $\begingroup$ Can we have $gcd(a,b)=gcd(c,d)=1$? $\endgroup$ – T.... Nov 16 '13 at 16:11

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