没有整数解的莫代尔曲线会有有理数解吗？

manthanein

假设方程\(y^2=x^3+p\)没有整数解，那么它会有有理数解吗？

设\(\D y=\frac{Y}{n}\)，\(\D x=\frac{X}{n}\)，代入后化为\(\D \frac{Y^2}{n^2}=\frac{X^3}{n^3}+p\)，去分母\(nY^2=X^3+n^3p\)，也即\((n^2Y)^2=(nX)^3+pn^6\)，也就是说，如果\(y^2=x^3+p\)没有有理数解，那么对于任意非零整数\(n\)，\(y^2=x^3+pn^6\)无整数解。
如何证明这一点？（或否定）

mathe

没有整数解但是有有理解的应该很常见

mathe

看 https://oeis.org/A054504 有对于p=6, 7, 11, 13, 14, 20, 21, 23, 29, 32, 34, 39, 42, 45, 46, 47, 51,...无整数解
计算对应曲线的阶
(22:18) gp > E=ellinit([0,7]);
(22:18) gp > ellanalyticrank(E)
%2 = [0, 3.0414172284278810424268528156464796264]
(22:18) gp > ellanalyticrank(ellinit([0,7]))
%3 = [0, 3.0414172284278810424268528156464796264]
(22:18) gp > ellanalyticrank(ellinit([0,14]))
%4 = [0, 2.7095947101363427554171013565399014510]
(22:19) gp > ellanalyticrank(ellinit([0,20]))
%5 = [0, 2.5532147367621155708875002997672877220]
(22:19) gp > ellanalyticrank(ellinit([0,21]))
%6 = [0, 2.5325369623438289744759653154434874276]
(22:19) gp > ellanalyticrank(ellinit([0,23]))
%7 = [0, 2.4944283611860673350734973257033976850]
(22:19) gp > ellanalyticrank(ellinit([0,29]))
%8 = [0, 2.3998974010004143986659319990931671331]
(22:19) gp > ellanalyticrank(ellinit([0,32]))
%9 = [0, 2.3608442970695375324491842692940779289]
(22:19) gp > ellanalyticrank(ellinit([0,34]))
%10 = [0, 2.3371101894880057595475144277220131475]
(22:19) gp > ellanalyticrank(ellinit([0,39]))
%11 = [1, 12.377052031405709570910216427916306508]
(22:19) gp > ellanalyticrank(ellinit([0,42]))
%12 = [0, 2.2562339333951459850961496144675544680]
(22:19) gp > ellanalyticrank(ellinit([0,45]))
%13 = [0, 2.2304385163122159011501697101803006765]
(22:19) gp > ellanalyticrank(ellinit([0,46]))
%14 = [1, 11.569116184827746560332118571869366616]
(22:19) gp > ellanalyticrank(ellinit([0,47]))
%15 = [1, 7.4994239040358623206273761551626204460]
(22:19) gp > ellanalyticrank(ellinit([0,51]))
%16 = [0, 2.1843923439202974271860166187381400929]
其中对于p=39,46,47对应的阶为1，说明对应的有理解存在，就是p=39,46,47都是反例

mathe

看链接https://oeis.org/A001014/a001014.txt
其中给出

E_+00039: r = 1   t = 1   #III =  1
          E(Q) = <(217/4, 3197/8)>
          R =   5.4183739683
           0 integral points
也就是说曲线$y^2=x^3+39$没有整数点但是有有理数点(217/4, 3197/8)作为生产元

xiaoshuchong

mathe 发表于 2018-5-27 06:47
看链接https://oeis.org/A001014/a001014.txt
其中给出

最有意思的例子是$y^2=x^3+7823$
没有整数解，最小有理点是
\[\begin{eqnarray*}
x&=&\frac{2263582143321421502100209233517777}{11981673410095561^2}\\
y&=&\frac{186398152584623305624837551485596770028144776655756}{11981673410095561^3}
\end{eqnarray*}\]