数论基础阅读笔记与习题解答之五:佩尔方程

《数论基础》第五章介绍了佩尔方程,以下是各节的习题解答:

5.1 边数与对角数

5.1.1 据题意有:

$$ \left( 1+\sqrt{2}\right)^{2} =1+2\sqrt{2} +2=3+2\sqrt{2} $$

进一步有:

$$ \left( x+y\sqrt{2}\right)\left( 1+\sqrt{2}\right) =x+x\sqrt{2} +y\sqrt{2} +2y=( x+2y) +( x+y)\sqrt{2} $$

5.1.2\(n=1\)时,有\(d_1=3,s_1=2\),设当\(n=k\)\((1+\sqrt{2})^{k+1}=d_k+s_k\sqrt{2}\),则当\(n=k+1\)时有:

$$ \begin{aligned} \left( 1+\sqrt{2}\right)^{k+2} & =\left( 1+\sqrt{2}\right)^{k+1}\left( 1+\sqrt{2}\right)\\ & =\left( d_{k} +s_{k}\sqrt{2}\right)\left( 1+\sqrt{2}\right)\\ & =d_{k} +d_{k}\sqrt{2} +s_{k}\sqrt{2} +2s_{k}\\ & =( d_{k} +2s_{k}) +( d_{k} +s_{k})\sqrt{2}\\ & =d_{k+1} +s_{k+1}\sqrt{2} \end{aligned} $$

也即\((1+\sqrt{2})^{n+1}=d_n+s_n\sqrt{2}\)

5.1.3 据题意有\(x^2-y^2=(x+y)(x-y)=1\),则有\(x+y=x-y=1\)或者\(x+y=x-y=-1\),对于前者可以得到\(x=1,y=0\),对于后者可以得到\(x=-1,y=0\),从而原方程有两组整数解。

5.1.4 据题意设\(n=d^2\),则\(x^2-ny^2=x^2-d^2y^2=(x+dy)(x-dy)=1\),则有:

$$ x+dy=x-dy=1\Rightarrow2x=2\Rightarrow x=1,y=0 $$

以及:

$$ x+dy=x-dy=-1\Rightarrow2x=-2\Rightarrow x=-1,y=0 $$

5.2 佩尔方程\(x^2-2y^2=1\)

5.2.1\((**)\)式因式分解并将第一与第三项,第二与第四项结合,易得:

$$ \begin{aligned} 1 & =\left( x_{1} +y_{1}\sqrt{2}\right)\left( x_{1} -y_{1}\sqrt{2}\right)\left( x_{2} +y_{2}\sqrt{2}\right)\left( x_{2} -y_{2}\sqrt{2}\right)\\ & =\left( x_{1} x_{2} +x_{1} y_{2}\sqrt{2} +x_{2} y_{1}\sqrt{2} +2y_{1} y_{2}\right)\left( x_{1} x_{2} -x_{1} y_{2}\sqrt{2} -x_{2} y_{1} +2y_{1} y_{2}\right)\\ & =\left( x_{1} x_{2} +2y_{1} y_{2} +( x_{1} y_{2} +x_{2} y_{1})\sqrt{2}\right)\left( x_{1} x_{2} +2y_{1} y_{2} -( x_{1} y_{2} +x_{2} y_{1})\sqrt{2}\right) \end{aligned} $$

5.2.2 将原式因式分解有\(x_3^2-2y_3^2=(x_3+y_3\sqrt{2})(x_3-y_3\sqrt{2})\),对照一下5.2.1的结果,易得:

$$ x_3=x_1x_2+2y_1y_2,y_3=x_1y_2+y_1x_2 $$

5.3 解的群

5.3.1 据题意有:

$$ \left( 3-2\sqrt{2}\right)^{k} =\left(\frac{1}{3+2\sqrt{2}}\right)^{k} =\frac{1}{\left( 3+2\sqrt{2}\right)^{k}} =\frac{1}{u_{k} +\sqrt{2} v_{k}} =\frac{u_{k} -\sqrt{2} v_{k}}{u_{k}^{2} -2v_{k}^{2}} =u_{k} -\sqrt{2} v_{k} $$

5.3.2 根据上题的结果有:

$$ \begin{cases} u_{k} +\sqrt{2} v_{k} =\left( 3+2\sqrt{2}\right)^{k}\\ u_{k} -\sqrt{2} v_{k} =\left( 3-2\sqrt{2}\right)^{k} \end{cases} \Rightarrow \begin{cases} 2u_{k} =\left( 3+2\sqrt{2}\right)^{k} +\left( 3-2\sqrt{2}\right)^{k}\\ 2\sqrt{2} v_{k} =\left( 3+2\sqrt{2}\right)^{k} -\left( 3-2\sqrt{2}\right)^{k} \end{cases} $$

则可得:

$$ u_{k} =\frac{1}{2}\left[\left( 3+2\sqrt{2}\right)^{k} +\left( 3-2\sqrt{2}\right)^{k}\right] ,v_{k} =\frac{1}{2\sqrt{2}}\left[\left( 3+2\sqrt{2}\right)^{k} -\left( 3-2\sqrt{2}\right)^{k}\right] $$

5.3.3 略。

5.4 广义佩尔方程与\(\mathbb{Z}[\sqrt{n}]\)

5.4.1\(\alpha,\beta\in \mathbb{Q}[n]\),则有\(\alpha=x_1+y_1\sqrt{n},\beta=x_2+y_2\sqrt{n}\),则有:

$$ \begin{aligned} \alpha +\beta & =x_{1} +y_{1}\sqrt{n} +x_{2} +y_{2}\sqrt{n} =( x_{1} +x_{2}) +( y_{1} +y_{2})\sqrt{n} \end{aligned} $$

显然\((x_1+x_2),(y_1+y_2)\in \mathbb{Q}\),则\(\mathbb{Q}[n]\)中元素对于加法是封闭的。另有:

$$ \begin{aligned} \alpha -\beta & =x_{1} +y_{1}\sqrt{n} -x_{2} -y_{2}\sqrt{n} =( x_{1} -x_{2}) +( y_{1} -y_{2})\sqrt{n} \end{aligned} $$

显然\((x_1-x_2),(y_1-y_2)\in \mathbb{Q}\),则\(\mathbb{Q}[n]\)中元素对于减法是封闭的。另有:

$$ \begin{aligned} \alpha \beta & =\left( x_{1} +y_{1}\sqrt{n}\right)\left( x_{2} +y_{2}\sqrt{n}\right) =( x_{1} x_{2} +ny_{1} y_{2}) +( x_{1} y_{2} +x_{2} y_{1})\sqrt{n} \end{aligned} $$

显然\((x_1x_2+ny_1y_2),(x_1y_2+x_2y_1)\in \mathbb{Q}\),则\(\mathbb{Q}[n]\)中元素对于乘法是封闭的。

5.4.2 据题意有:

$$ \frac{1}{x+y\sqrt{n}} =\frac{1}{x+y\sqrt{n}} \cdot \frac{x-y\sqrt{n}}{x-y\sqrt{n}} =\frac{x-y\sqrt{n}}{x^{2} -ny^{2}} =\frac{x}{x^{2} -ny^{2}} +\frac{-y}{x^{2} -ny^{2}}\sqrt{n} $$

\(x'=\frac{x}{x^{2} -ny^{2}} ,y'=\frac{-y}{x^{2} -ny^{2}}\),则显然有\(x',y'\in \mathbb{Q}\)。则对任意\(\alpha,\beta\in \mathbb{Q}[n]\),则\(\frac{\alpha }{\beta } =\alpha \times \frac{1}{\beta }\),上面我们已经\(\mathbb{Q}[n]\)内的乘法运算是封闭的,而\(1/\beta\in \mathbb{Q}\),则\(\alpha/\beta\in\mathbb{Q}\),即\(\mathbb{Q}[n]\)下除法运算对于\(\beta\ne0\)是封闭的。

5.4.3 据题意有:

$$ \begin{array}{ l } ( x_{1} x_{2} +ny_{1} y_{2})^{2} -n( x_{1} y_{2} +y_{1} x_{2})^{2}\\ =x_{1}^{2} x_{2}^{2} +2nx_{1} x_{2} y_{1} y_{2} +n^{2} y_{1}^{2} y_{2}^{2} -nx_{1}^{2} y_{2}^{2} -2nx_{1} x_{2} y_{1} y_{2} -ny_{1}^{2} x_{2}^{2}\\ =x_{1}^{2}\left( x_{2}^{2} -ny_{2}^{2}\right) -ny_{1}^{2}\left( x_{2}^{2} -ny_{2}^{2}\right)\\ =\left( x_{1}^{2} -ny_{1}^{2}\right)\left( x_{2}^{2} -ny_{2}^{2}\right)\\ =k_{1} k_{2} \end{array} $$

5.4.4 易知方程\(x^2-17y^2=-1\)的最小正整数解是\((4,1)\),则根据5.4.3的结论可以推出如果\((x,y)\)是满足方程\(x^2-dy^2=-1\)的解,则\((x^2+dy^2,2xy)\)就是满足方程\(x^2-dy^2=1\)的解,则可以得到\((4^2+17\times1,2\times4\times1)=(33,8)\)是方程\(x^2-dy^2=1\)的一个解。

5.4.5 易知方程\(x^2-37y^2=-1\)的最小正整数解是\((6,1)\),则\((6^2+37,2\times6)=(73,12)\)应该是方程\(x^2-37y^2=1\)的一个解。

5.6 二次型

5.6.1 题目中二次型对应的矩阵为:

$$ \begin{bmatrix} A & B/2\\ B/2 & C \end{bmatrix} =\begin{bmatrix} 13 & 8\\ 5 & 5 \end{bmatrix} $$

则其行列式为\(D=13\times5-8^2=1\)。另根据\(M\)的值,有:

$$ ( 2x+y)^{2} +( 3x+2y)^{2} =4x^{2} +4xy+y^{2} +9x^{2} +12xy+4y^{2} =13x^{2} +16xy+5y^{2} $$

所以在\(M\)下原二次型与\(x^2+y^2\)是等价的。

5.6.2 二次型\(2x^2+2xy+3y^2\)对应矩阵的行列式为:

$$ det\left(\begin{matrix} 2 & 1\\ 1 & 3 \end{matrix}\right) =6-1=5 $$

另二次型\(x^2+5y^2\)对应矩阵的行列式为:

$$ det\left(\begin{matrix} 1 & 0\\ 0 & 5 \end{matrix}\right) =5-0=5 $$

两个二次型的行列式是相等的。显然当\(x=y=1\)时,\(2x^2+2xy+3y^2=7\)。如果\(x^2+5y^2=7\),则有\(x^2=7-5y^2\ge0\Rightarrow-1.183\le y\le1.183\),则\(y\)可取的整数为\(-1,0,1\),则有:

所以\(x^2+5y^2\)不可能取得\(7\),所以\(2x^2+2xy+3y^2\)不可能等价于\(x^2+5y^2\)

5.6.3 易知\(x^2\bmod20\)的取值可有\(0,1,4,5,9,16\),则\(5y^2\bmod20\)取值有\(0,5\),则可得:

$$ x^2+5y^2\equiv(0,1,4,5,6,9,10,14,16)\pmod{20} $$

显然取值中不包含\(3,7\)