問題の解答
検索用コード（LaTeX）
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% 例題A4.2.10：方程式の整数解8（One More）★★★★
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$3x^2+4xy-4y^2+4x-16y-20=0$を満たす整数の組$(x,y)$を求めよ．
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% 例題A4.2.10の解答（One More）★★★★
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\[
3x^2+4xy-4y^2=(3x-2y)(x+2y)
\]
と因数分解できるので，定数$p,q$を用いて$(3x-2y+p)(x+2y+q)$を展開し，与えられた式の左辺と比較する．
\begin{align*}
(3x-2y+p)(x+2y+q)&=(3x-2y)(x+2y)+q(3x-2y)+p(x+2y)+pq\\
&=3x^2+4xy-4y^2+(p+3q)x+(2p-2q)y+pq
\end{align*}
したがって，与えられた式と$x,y$の項の係数を比較すると，
\[
\{
\begin{array}{l}
p+3q=4\\
2p-2q=-16
\end{array}
\]
これを解くと，$p=-5,q=3$
ゆえに，
\[
(3x-2y-5)(x+2y+3)=3x^2+4xy-4y^2+4x-16y-15
\]
したがって，与えられた式は，
\[
3x^2+4xy-4y^2+4x-16y-15-5=0
\]
整理すると，
\[
(3x-2y-5)(x+2y+3)=5
\]
ゆえに，
\[
(3x-2y-5,x+2y+3)=(1,5),(5,1),(-1,-5),(-5,-1)
\]
これを解いて，
\[
(x,y)=(2,0),(2,-2),(-1,-\frac{7}{2}),(-1,-\frac{3}{2})
\]
よって，$x,y$は整数より，$(x,y)=(2,0),(2,-2)$
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% 問題A4.2.10：方程式の整数解8（One More）★★★★
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$2x^2-7xy+3y^2+8x-9y-5=0$を満たす整数の組$(x,y)$を求めよ．
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% 問題A4.2.10の解答（One More）★★★★
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\[
2x^2-7xy+3y^2=(2x-y)(x-3y)
\]
と因数分解できるので，定数$p,q$を用いて$(2x-y+p)(x-3y+q)$を展開し，与えられた式の左辺と比較する．
\begin{align*}
(2x-y+p)(x-3y+q)&=(2x-y)(x-3y)+q(2x-y)+p(x-3y)+pq\\
&=2x^2-7xy+3y^2+(p+2q)x-(3p+q)y+pq
\end{align*}
したがって，与えられた式と$x,y$の項の係数を比較すると，
\[
\{
\begin{array}{l}
p+2q=8\\
-3p-q=-9
\end{array}
\]
これを解くと，$p=2,q=3$
ゆえに，
\[
(2x-y+2)(x-3y+3)=2x^2-7xy+3y^2+8x-9y+6
\]
したがって，与えられた式は，
\[
2x^2-7xy+3y^2+8x-9y+6-11=0
\]
整理すると，
\[
(2x-y+2)(x-3y+3)=11
\]
ゆえに，
\[
(2x-y+2,x-3y+3)=(1,11),(11,1),(-1,-11),(-11,-1)
\]
これを解いて，
\[
(x,y)=(-\frac{11}{5},-\frac{17}{5}),(\frac{29}{5},\frac{13}{5}),(1,5),(-7,-1)
\]
よって，$x,y$は整数より，$(x,y)=(1,5),(-7,-1)$
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