Showing $a^2-359b^2=5$ has no solutions
问题内容
I am trying to prove that the prime ideals above 5 in $K=\mathbb Q(\sqrt{359})$ are non-principal. I calculated the splitting to be $5O_K = (5,\sqrt{359} + 2)(5,\sqrt{359} + 3)$. If either of the ideals were principal their generator would have norm $\pm 5$.
Showing that $a^2-359b^2 = -5$ has no solutions worked fine by considering it $\text{mod } 8$, but I am at a loss trying to show $a^2-359b^2 =5$ has no integral solutions.
回答 (1)
The generalized Pell equation to be solved is $$ x^2 - 359y^2=5\ , $$ and note that $5$ is in the interval between $0$ and $\sqrt{359}$, so a theorem of Legendre (on the approximation of the algebraic number $\sqrt{359}$ by rational numbers for the algebraic degree two, similar results being obtained in higher algebraic degrees by Thue, Siegel, Roth) insures the property from the wiki page:
If $x$ and $y$ are positive integer solutions to the Pell's equation with $| \displaystyle |N| < \sqrt {n}$, then $x / y$ is a convergent to the continued fraction of $\sqrt {n}$. (See the #35 on the wiki page for a reference.)
Using the fundamental unit $U = 360+19\sqrt{359}$ and its powers we can moreover make $x+y\sqrt{359}$ become small, and the same wiki page gives for $x$ the upper bound, $|x|\le \frac 12\sqrt{|N|}(\sqrt{|U|+1})$.
In our case $N=5=\sqrt{25}$ is smaller $\sqrt{359}$, and the first few convergents $x/y$, and norms of $x+y\sqrt{359}$ are: $$ \begin{array}{|c|c|c|} \hline x & y & x^2-359y^2 \\\hline 18 & 1 & -35\\\hline 19 & 1 & 2\\\hline \color{gray}{341} & \color{gray}{18} & \color{gray}{-35}\\\hline \color{gray}{360} & \color{gray}{19} & \color{gray}{1}\\\hline \end{array} $$ And we may stop after the first two lines, the third one is beyond the $x$-upper bound. So only $1,2$ can be realized as norms among the numbers up to $18$.