問題の解答
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% 例題A4.2.11：方程式の整数解9（One More）★★★★
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方程式$x^2+2xy+5y^2+4x-12y+11=0$を満たす整数の組$(x,y)$をすべて求めよ．
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% 例題A4.2.11の解答（One More）★★★★
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$x^2+2xy+5y^2+4x-12y+11=0$を$x$について整理すると，
\[
x^2+2(y+2)x+(5y^2-12y+11)=0\cdots(\mathrm{i})
\]
2次方程式の判別式を$D$とすると，
\begin{align*}
\frac{D}{4}&=(y+2)^2-1\cdot(5y^2-12y+11)\\
&=y^2+4y+4-(5y^2-12y+11)\\
&=-4y^2+16y-7
\end{align*}
(i) の解が実数となるから，$D\geqq 0$
したがって，$-4y^2+16y-7\geqq 0$より，$(2y-1)(2y-7)\leqq 0$
ゆえに，$\frac{1}{2}\leqq y\leqq\frac{7}{2}$
$y$は整数であるから，$y=1,2,3$
（ア）$y=1$のとき，(i)より，$x^2+6x+4=0$
これを解くと，$x=-3\pm{\sqrt{5}}$となり，不適である．
（イ）$y=2$のとき，(i)より，$x^2+8x+7=0$
これを解くと，$x=-7,-1$
（ウ）$y=3$のとき，(i)より，$x^2+10x+20=0$
これを解くと，$x=-5\pm\sqrt{5}$となり，不適である．
よって，（ア）〜（ウ）より，$(x,y)=(-7,2),(-1,2)$
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% 問題A4.2.11：方程式の整数解9（One More）★★★★
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方程式$x^2-2xy+2y^2-2x-2y+4=0$を満たす整数の組$(x,y)$をすべて求めよ．
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% 問題A4.2.11の解答（One More）★★★★
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$x^2-2xy+2y^2-2x-2y+4=0$を$x$について整理すると，
\[
x^2-2(y+1)x+(2y^2-2y+4)=0\cdots(\mathrm{i})
\]
2次方程式の判別式を$D$とすると，
\[
\frac{D}{4}=(y+1)^2-1\cdot(2y^2-2y+4)=y^2+2y+1-(2y^2-2y+4)=-y^2+4y-3
\]
(i) の解が実数となるから，$\frac{D}{4}\geqq 0$
したがって，$-y^2+4y-3\geqq 0$より，$-(y-1)(y-3)\geqq 0$
ゆえに，$1\leqq y\leqq 3$
$y$は整数であるから，$y=1,2,3$
（ア）$y=1$のとき，(i)より，$x^2-4x+4=0$
これを解くと，$x=2$
（イ）$y=2$のとき，(i)より，$x^2-6x+8=0$
これを解くと，$x=4,2$
（ウ）$y=3$のとき，(i)より，$x^2-8x+16=0$
これを解くと，$x=4$
よって，（ア）〜（ウ）より，$(x,y)=(2,1),(4,2),(2,2),(4,3)$
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