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2元1次不定方程式( indefinite equation?)<2/4>「2022年度 大学入学共通テスト 本試験|数学Ⅰ第4問」(配点20点)WolframAlphaで(合同式も)

Last updated at Posted at 2022-03-30

つづき

(20220330)これも考察?を使っていません。
(20220404)これも考察?を使っていません。後半にWolframAlphaの合同式(mod)でやってみました。
オリジナル

WolframAlphaで

ア,イウ n = 0, x = 1, y = 39

結果
x = 16 n + 1 , n>=1 , n element Z

エオ,カキク n = 1, x = 17, y = 664

整数解

ケ x = 8

結果

コ x = 5 , m = 39

結果
x = 32 n + 29 , y = 3125 n + 2832 , n element Z

結果
x = 32 n + 29 , n>=3 , n element Z

サシス,セソタチツ n = 3, x = 125, y = 12207

結果
x = 32 n + 19 , y = 161051 n + 95624 , n element Z

テト,ナニヌネノ n = 0, x = 19, y = 95624

wolfram alpha 合同式でやってみました。

roundup かっこ悪いですね。いい方法がありますか?

(1) ア,イ,ウ

5^4x-2^4y=1

General solution
x = 16 n + 1 and n element Z

Result

n = 0, x = 1, y = 39

(1) エオ,カキク

整数解の例
n = 1 , x = 17

Solution
{625 x - 16 y = 1, x = 17}

(2)ケ,コ

625^2=5^x

整数解
x = 8

Integer solution
m = 39, x = 5

(3) サシス,セソタチツ

5^5x-2^5y=1

General solution
x = 32 n + 29 and n element Z

Solution
n>=71/32

roundup(71/32)

Exact result
125

Solution
x = 125, y = 12207

(4) テト,ナニヌネノ

11^5x-2^5y=1

General solution

x = 32 n + 19 and n element Z

Solution
x = 19, y = 95624

参考

Congruence(合同:google翻訳より[UK]コングルアンス[US]カングルアンス)

つづく <3/4> へ

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