【数学A】例題4.2.8：方程式の整数解6（One More）★★★

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% 例題A4.2.8：方程式の整数解6（One More）★★★
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(1) $x^2-y^2=99$を満たす自然数の組$(x,y)$をすべて求めよ．

(2) $\sqrt{n^2-35}$が自然数となるような自然数$n$をすべて求めよ．

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% 例題A4.2.8の解答（One More）★★★
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(1) $x^2-y^2=99$より，$(x-y)(x+y)=99$

ここで，$x,y$は自然数であり，$x^2-y^2>0$より，$x>y$であるから，$x-y,x+y$も自然数であり，

\[
x-y<x+y
\]

よって，$(x-y,x+y)=(1,99),(3,33),(9,11)$

(i) $x-y=1,x+y=99$のとき，$(x,y)=(50,49)$

(ii) $x-y=3,x+y=33$のとき，$(x,y)=(18,15)$

(iii) $x-y=9,x+y=11$のとき，$(x,y)=(10,1)$

(i)〜(iii)より，求める自然数の組は，

\[
(x,y)=(50,49),(18,15),(10,1)
\]

(2) $\sqrt{n^2-35}=m$（$m$は自然数）とおく．

両辺を2乗すると，$n^2-35=m^2$

したがって，$n^2-m^2=35$より，$(n-m)(n+m)=35$

ここで，$n,m$は自然数であり，$n^2-m^2>0$より，$n>m$であるから，$n-m,n+m$も自然数であり，

\[
n-m<n+m
\]

ゆえに，$(n-m,n+m)=(1,35),(5,7)$

(i) $n-m=1,n+m=35$のとき，$(n,m)=(18,17)$

(ii) $n-m=5,n+m=7$のとき，$(n,m)=(6,1)$

よって，(i)，(ii)より，$n=6,18$

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% 問題A4.2.8：方程式の整数解6（One More）★★★
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(1) $x^2-y^2=45$を満たす自然数の組$(x,y)$をすべて求めよ．

(2) $\sqrt{n^2-63}$が自然数となるような自然数$n$をすべて求めよ．

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% 問題A4.2.8の解答（One More）★★★
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(1) $x^2-y^2=45$より，$(x-y)(x+y)=45$

ここで，$x,y$は自然数であり，$x^2-y^2>0$より，$x>y$であるから，$x-y,x+y$も自然数であり，

\[
x-y<x+y
\]

よって，$(x-y,x+y)=(1,45),(3,15),(5,9)$

(i) $x-y=1,x+y=45$のとき，$(x,y)=(23,22)$

(ii) $x-y=3,x+y=15$のとき，$(x,y)=(9,6)$

(iii) $x-y=5,x+y=9$のとき，$(x,y)=(7,2)$

(i)〜(iii)より，求める自然数の組は，

\[
(x,y)=(23,22),(9,6),(7,2)
\]

(2) $\sqrt{n^2-63}=m$（$m$は自然数）とおく．

両辺を2乗すると，$n^2-63=m^2$

したがって，$n^2-m^2=63$より，$(n-m)(n+m)=63$

ここで，$n,m$は自然数であり，$n^2-m^2>0$より，$n>m$であるから，$n-m,n+m$も自然数であり，

\[
n-m<n+m
\]

ゆえに，$(n-m,n+m)=(1,63),(3,21),(7,9)$

(i) $n-m=1,n+m=63$のとき，$(n,m)=(32,31)$

(ii) $n-m=3,n+m=21$のとき，$(n,m)=(12,9)$

(iii) $n-m=7,n+m=9$のとき，$(n,m)=(8,1)$

よって，(i)〜(iii)より，$n=8,12,32$

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% 例題A4.2.8の側注：方程式の整数解6（One More）★★★
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% 側注1
連立方程式を解くことで，$(x,y)$を求めることができる．

% 側注2
$y>0$より，$x-y$と$x+y$の大小が定まる．

% 側注3
$x-y,x+y$はともに$99=3^2\cdot 11$の正の約数である．

% 側注4
$n-m,n+m$はともに$35=5\cdot 7$の正の約数である．

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% 問題A4.2.8の側注：方程式の整数解6（One More）★★★
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% 側注1
連立方程式を解くことで，$(x,y)$を求めることができる．

% 側注2
$y>0$より，$x-y$と$x+y$の大小が定まる．

% 側注3
$x-y,x+y$はともに$45=3^2\cdot 5$の正の約数である．

% 側注4
$n-m,n+m$はともに$63=3^2\cdot 7$の正の約数である．

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% 例題A4.2.8の考え方・補足など：方程式の整数解6（One More）★★★
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% 考え方
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$A,B$が整数のとき，例えば，$AB=3$であれば，$(A,B)=(3,1),(1,3),(-1,-3),(-3,-1)$と3の約数を考えることで$A,B$の値を定めることができる．

このことを利用するために，$AB=(\text{整数})$の形に式変形することを考える．(2)は，$\sqrt{n^2-35}=m$（$m$は自然数）とおき，両辺を2乗して，$n^2-m^2=35$と式変形すればよい．