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I am looking for infinite set of Diophantine solutions.

  1. Suppose we require $$0<\min(a,d)<\max(a,d)<\min(b,c)<\max(b,c)\leq\sqrt 2\min(a,d)$$ $$a,b,c,d\in\mathbb Z$$ then can we still find solutions to $$ab-cd=1?$$

  2. Is it possible to do this if only $$a,b,c,d\in\mathbb Z$$ $$0<\min(a,d)<\max(a,d)<\min(b,c)<\max(b,c)$$ $$ad<ab+cd\leq2ad\leq2bc$$ holds?

  3. If not what is the closest permissible?

What is a good bound?

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  • $\begingroup$ Note 2. is not necessarily weaker and is a different statement if $ab+cd<bc$ does not hold. $\endgroup$
    – VS.
    Commented Aug 11, 2020 at 22:59

1 Answer 1

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It is possible with any constant $\lambda>1$ on the place of $\sqrt{2}$. Take $a=t^2$, $d=t^2+t-1$, $b=t^2+2t+1$, $c=t^2+t+1$ for large $t$.

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  • $\begingroup$ nicely done!... $\endgroup$ Commented Aug 11, 2020 at 22:25
  • $\begingroup$ @FedorPetrov Is it also possible to do 2. or at least do it with a different example? In your example $ab+cd<bc$ does not hold. $\endgroup$
    – VS.
    Commented Aug 11, 2020 at 22:56
  • $\begingroup$ @VS. I am confused, what exact conditions do you require now? I thought that 2. is $ad<ab+cd\leqslant 2ad$ which clearly holds for this example. $\endgroup$ Commented Aug 11, 2020 at 23:04
  • $\begingroup$ 2. is impossible. Assume $b=d+\delta_1$ and $c=a+\delta_2$. Then $ab+cd=2ad+a\delta_1+d\delta_2$ and since $\max(b,c)\leq\min(a,d)$ we have $\delta_1,\delta_2>0$. $\endgroup$
    – VS.
    Commented Aug 12, 2020 at 0:32

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