\(ad -bc = 1 , \ a \gt 0\) を満たす整数 \(a , b , c , d\) を考える. 行列 \[\begin{align} A & = \left( \begin{array}{cc} 6 & 10 \\ 10 & 17 \end{array} \right) , \quad B = \left( \begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array} \right) , \\ M & = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) , \quad N = \left( \begin{array}{cc} a & c \\ b & d \end{array} \right) \ . \end{align}\] が \(NA = BM^{-1}\) を満たすとき, 以下の各問いに答えよ. ただし, \(M^{-1}\) は \(M\) の逆行列を表す.
(1) \(6a^2+20ac+17c^2\) の値を求めよ.
(2) \(2a^2+b^2\) の値を求めよ.
(3) \(a , b , c , d\) の値を求めよ.
(4) \(6x^2+20xy+17y^2 = 59\) を満たす実数 \(x , y\) に対して \[ \left\{ \begin{array}{l} X = dx-by \\ Y = -cx+ay \end{array} \right. \ . \] とおくとき, \(X^2+2Y^2\) の値を求めよ.
(5) \(6x^2+20xy+17y^2 = 59\) を満たす整数の組 \((x,y)\) をすべて求めよ.
【 解 答 】
(1)
条件より, \(NAM = B\) . \[\begin{align} NAM & = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{cc} 6 & 10 \\ 10 & 17 \end{array} \right) \left( \begin{array}{cc} a & c \\ b & d \end{array} \right) \\ & = \left( \begin{array}{cc} 6a+10b & 10a+17b \\ 6c+10d & 10c+17d \end{array} \right) \left( \begin{array}{cc} a & c \\ b & d \end{array} \right) \\ & = \left( \begin{array}{cc} 6a^2+20ab+17b^2 & 6ac+10(ad+bc)+17bd \\ 6ac+10(ad+bc)+17bd & 6c^2+20cd+17d^2 \end{array} \right) \ . \end{align}\] よって, \((1,1)\) 成分を比較して \[ 6a^2+20ab+17b^2 = \underline{1} \quad ... [1] \ . \]
(2)
\(ad -bc = 1\) より \[ M^{-1} = \left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right) , \ N^{-1} = \left( \begin{array}{cc} d & -c \\ -b & a \end{array} \right) \ . \] また, 条件より, \(A = N^{-1}BM^{-1}\) . \[\begin{align} N^{-1}BM^{-1} & = \left( \begin{array}{cc} d & -c \\ -b & a \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array} \right) \left( \begin{array}{cc} d & -c \\ -b & a \end{array} \right) \\ & = \left( \begin{array}{cc} d & -2c \\ -b & 2a \end{array} \right) \left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right) \\ & = \left( \begin{array}{cc} 2c^2+d^2 & -bd-2ac \\ -bd-2ac & 2a^2+b^2 \end{array} \right) \ . \end{align}\] よって, \((2,2)\) 成分を比較して \[ 2a^2+b^2 = \underline{17} \quad ... [2] \ . \]
(3)
(1) (2) について, 他の成分についても比較すれば
\[\begin{align}
6c^2+20cd+17d^2 & = 2 \quad ... [3] , \\
6ac+10(ad+bc)+17bd & = 0 \quad ... [4] , \\
2c^2+d^2 & = 6 \quad ... [5] , \\
-bd -2ac & = 10 \quad ... [6] \ .
\end{align}\]
\(a , b, c , d\) はすべて整数なので, [5] より
\[
(c,d) = ( \pm 1 , \pm 2 ) \ .
\]
同様に考えて, \(a \gt 0\) と [2] より
\[
(a,b) = ( 2 , \pm 3 ) \ .
\]
このうち, \(ad -bc = 1\) をみたす組は, \(|ad| = 4\) , \(|bc| = 3\) なので
\[
(a,b,c,d) = ( 2 , \pm 3 , \pm 1 , 2 ) , ( 2 , \mp 3 , \mp 1 , 2 ) \quad ( \text{複号同順} ) \ .
\]
さらに, [6] をみたす組は
\[
(a,b,c,d) = ( 2 , -3 , -1 , 2 ) \ .
\]
これは [3] [4] もみたしている.
よって
\[
(a,b,c,d) =\underline{( 2 , -3 , -1 , 2 )} \ .
\]
(4)
条件より \[ \left( \begin{array}{c} X \\ Y \end{array} \right) = M^{-1} \left( \begin{array}{c} x \\ y \end{array} \right) \ . \] なので \[ \left( \begin{array}{c} x \\ y \end{array} \right) = M \left( \begin{array}{c} X \\ Y \end{array} \right) = \left( \begin{array}{c} 2X-3Y \\ -X+2Y \end{array} \right) \quad ... [7] \ . \] これを \(6x^2+20xy+17y^2 = 59\) に代入すれば \[\begin{align} 6 (2X-3Y)^2 +20(2X-3Y)(-X+2Y) +17(-X+2Y)^2 & = 59 \\ 24X^2-72XY+54Y^2 -40X^2+140XY \qquad & \\ -120Y^2 +17X^2-68XY+68Y^2 & = 59 \\ \text{∴} \quad X^2+2Y^2 = \underline{59} \quad ... [8] & \ . \end{align}\]
(5)
\(x , y\) が整数ならば, [7] より \(X , Y\) も整数である.
[8] より, \(0 \leqq Y^2 \leqq \dfrac{59}{2}\) なので, [8] の解は
\[
(X,Y) = ( \pm 3 , \pm 5 ) \ .
\]
よって, [7] より求める解は
\[
(x,y) = \underline{( \pm 9 , \mp 7 ) , ( \pm 21 , \mp 13 )} \quad ( \text{複号同順} ) \ .
\]