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[数论] $y^2+54=x^3$

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hbghlyj posted 2024-4-11 07:50 |Read mode
section.12 最后的Remarks 说道,
try using similar techniques to find all solutions to $y^2+54=x^3$. Unlike the example we went over in detail, this equation does have some solutions.
WolframAlpha:Solve[y^2 + 54 == x^3, {x, y}, Integers]
$$x = 7,\quad y = ± 17$$

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original poster hbghlyj posted 2024-4-11 07:56
$(y+\sqrt{-54})(y-\sqrt{-54})=x^3$
那么它们生成的理想也满足方程
$(y+\sqrt{-54})(y-\sqrt{-54})=(x^3)$
要证明$(y+\sqrt{-54})$与$(y-\sqrt{-54})$互质。

证明:??

继续解原方程:
由$\mathbb Z[\sqrt{-6}]$ is Dedekind domain 推出$(y+\sqrt{-54})={\frak a}^3$,$\frak a$ 是一个理想,
NumberFieldClassNumber[Sqrt[-6]]=2
$[{\frak a}^3]=[(1)]$,2和3互质,所以 $\frak a$ 是主理想,令 ${\frak a}=(a+b \sqrt{-6}),a,b\inZ$.
$$y+\sqrt{-54}=u(a+b \sqrt{-6})^3$$
$u\in U(\mathbb Z[\sqrt{-6}])=\{\pm1\}$,因为$(-1)^3=-1$,不妨设$u=1$.
$$y+\sqrt{-54}=(a+b \sqrt{-6})^3$$

$$y+3\sqrt{-6}=(a+b \sqrt{-6})^3$$
对比$\sqrt{-6}$的系数,$\cases{y=a(a^2-18b^2)\\3=b(3a^2-6b^2)\implies1=b(a^2-2b^2)\implies b=-1,a=\pm1}$
代回去就得到 $x = 7,y = ± 17$

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