1

The following lines are from Schinzels paper "integer points on conics"

Let $ax^2 + bxy + cy^2 + dx + ey +f = 0$ be a non-degenerate conic over integers.

Setting $X= \delta x+be-2cd$, $Y=\delta y+bd-2ae$, where $\delta=b^2-4ac$, we get $$aX^2 +bXY+cY^2=4\delta \Delta \ \ \ (2) $$ for some integer $\Delta$.

Let $(u_0,v_0)$ be the least positive solution for $u^2-4\delta^3v^2=4$ and set $\varepsilon=(\frac{u_0+2\delta \sqrt{\delta}v_0}{2})^2 $.

If $X_0,Y_0$ is a solution of (2) and satisfies the congruences $$ X_0 \equiv be-2cd \bmod \delta,\ \ \ Y_0 \equiv bd-2ae \bmod \delta$$ then also every solution $X,Y$ of $$ 2aX +(b+\sqrt{\delta})Y = (2aX_0 +(b+\sqrt{\delta})Y_0) \varepsilon ^n$$satisfies these congrunces. Therefore we can assume that $$ 4 \sqrt{|a\delta\Delta|} \leq |2aX_0 +(b+\sqrt{\delta})Y_0| \leq 4 \sqrt{|a\delta\Delta|}\varepsilon $$ Question: Can somebody explain me why we can assume this?

paris
  • 11
  • do you have a link to the article? I do not see it online, although I can get a scan from the campus library in a few days – Will Jagy Jul 27 '16 at 16:59
  • hmmm. No, they do not have that volume, or the later volume with the errata – Will Jagy Jul 27 '16 at 18:45
  • Dear Will, thank you very very much for your time! I didn't recieve any notification for your answer, so I am replying quite late... This is actually the erata. I missed to write that epsilon>1, so if you concider |2aX_0+(b+δ√)Y_0|>4|aδΔ| then multiplying (2aX0+(b+δ√)Y0) by the conjugate of epsilon which is <1 you get a smaller solution which is a contrudiction to the assumption... – paris Sep 05 '16 at 10:02
  • Of course if you want to see the paper i can scan it and send it you... Thanks again! – paris Sep 05 '16 at 10:05

0 Answers0