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数学 高校生

この整数解のひとつを求めるのはどうすればよいのですか?あと、求め方は解答には書かなくていいんでしょうか?教えてください!

ーーなーーっ ーーの〇〇① ! 次の方程式の整数解をすべて求めよ。 ⑦ 3メークニュ (2⑫) 22x+37y=2 W 5 | スル9 本.革121 | 上 ] | @gasr@加orprron | 1 次不定方程式 <++ ゥyニc の整数解 1 組の解 (ヵ, o) を見つけて g(ーヵ)十6①ーの)三0 ……g | (3) 係数が小さいから, 1 組の解が見つけやすい。 (2) 係数が大きいから, 1 組の解が見つけにくい。そこで, 基本例題 121 のよぅに ① 2x+か1 の整数解 ニカ,ャニg を互除法を用いて求める。 ② の2g三1 から, 両辺にcを掛けて go(cの)士が(cg)三c の手順で進める。最後の式と ox十jーc から g(テメーcの)十6(ッーcog)=0 (CE ⑰ 3テー7タニ1 …… @ ィー5, 2 は, ① の整数解の 1 つである。 よう:C。 3-5一7・2ニ1 ……@② ⑦④-②から 3(ー5ー7①ー2)=0 すなわち 3ー5)=7⑦ー2) …… ③⑨ 3と7は互い あるから, ③より ィー5ニ7 ッー2ニ3ん (んは整数) したがって, ① のすべての整数解は ャニー7ん5, ッニ3ん十2 (は整数) 22*ナ37y=2 …… @① =ー5, ニー3 は, 22zす37yー1 の整数解の 1 つである。 って 22・(5)37・3=1 辺に 2 を掛けると 22・(一10)十37.6=2 ……② =のから| 22(ヶ上10)十 ?

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