\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=2\)

王守恩

王守恩 发表于 2020-12-17 12:43
举个例子。
\(\sqrt[29]{a+b\sqrt{30}}+\sqrt[29]{a-b\sqrt{30}}=2\ \ a=?\ \ b=?\)

给出\(\ \ \ \sqrt[2k+1]{a+b\sqrt{c}}+\sqrt[2K+1]{a-b\sqrt{c}}=2\ \ \ \)通项公式，请各位网友批评!
在这里，\(a, b\ \)是正整数(未知数)。c, k 是已知数\(\ \ c=1, 2, 3,...\ k=1, 2, 3,...\ \ \)

从简单说起
1，c=1
a=LinearRecurrence[{4,}, {1,}, k+1]
b=LinearRecurrence[{4,}, {1,}, k+1]
1，c=1
a=LinearRecurrence[{5, -4}, {1, 4}, k+1]
b=LinearRecurrence[{5, -4}, {1, 4}, k+1]

2，c=2
a=LinearRecurrence[{6, -1}, {1, 7}, k+1]
b=LinearRecurrence[{6, -1}, {1, 5}, k+1]

3，c=3
a=LinearRecurrence[{8, -4}, {1, 10}, k+1]
b=LinearRecurrence[{8, -4}, {1, 06}, k+1]

4，c=4
a=LinearRecurrence[{10, -9}, {1, 13}, k+1]
b=LinearRecurrence[{10, -9}, {1, 07}, k+1]

5，c=5
a=LinearRecurrence[{12, -16}, {1, 16}, k+1]
b=LinearRecurrence[{12, -16}, {1, 08}, k+1]

6，c=6
a=LinearRecurrence[{14, -25}, {1, 19}, k+1]
b=LinearRecurrence[{14, -25}, {1, 09}, k+1]

7，.......

8，c=c
a=LinearRecurrence[{2(c+1), -(c-1)^2}, {1, 3c+1}, k+1]
b=LinearRecurrence[{2(c+1), -(c-1)^2}, {1,  c+3}, k+1]

注：别小看这些数字串，可是在 OIES 找不到的。

王守恩

elim 发表于 2020-12-14 21:10
原方程\(\implies(2a-(n/m)^3)^3=27(n/m)^3（b^2c-a^2）\)
\(\implies(2am^3-n^3)^3=27n^3m^6(b^2c-a^2) ...

还有简单的！！！

给出\(\ \ \ \sqrt[2k+1]{a+b\sqrt{c}}+\sqrt[2K+1]{a-b\sqrt{c}}=2\ \ \ \)通项公式，请各位网友批评!

在这里，\(a, b\ \)是正整数(未知数)。c, k 是已知数\(\ \ c=1, 2, 3,...\ k=1, 2, 3,...\ \ \)

\(\  a=\frac{(1+\sqrt{c})^{2k+1}+(1-\sqrt{c})^{2k+1}}{2}\)

\(\  b=\frac{(1+\sqrt{c})^{2k+1}-(1-\sqrt{c})^{2k+1}}{2\sqrt{c}}\)