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

王守恩

相似的话题！若\(a, b, c\ \)是正整数。
试证：\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=\frac{1}{2}\ \)无解。

elim

王守恩 发表于 2020-12-13 17:07
相似的话题！若\(a, b, c\ \)是正整数。
试证：\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=\frac{1}{ ...

容易证明, 对\(\,1< n\in\mathbb{N},\)
\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}={\large\frac{1}{n}}\;\;(a,b,c\in\mathbb{N}^+)\)
无解.

王守恩

elim 发表于 2020-12-14 10:05
容易证明, 对\(\,1< n\in\mathbb{N},\)
\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}={\large\frac{1 ...

往前走！若\(a, b, c\ \)是正整数。
试证：\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=\frac{n}{m}\ \)无解。
\(n=1, 2, 3, 4, 5, .....\ \ m=2, 3, 4, 5, .....\)
\(只要n, m是互质数，则\frac{n}{m}无解。\)

elim

只要\(n,m\)互素且\(m>1\), 则 \(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=\large\frac{n}{m}\) 无整数解。

elim

elim 发表于 2020-12-14 05:46
只要\(n,m\)互素且\(m>1\), 则 \(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=\large\frac{n}{m}\) 无整 ...

原方程\(\implies(2a-(n/m)^3)^3=27(n/m)^3（b^2c-a^2）\)
\(\implies(2am^3-n^3)^3=27n^3m^6(b^2c-a^2)\). 设素数 \(p\mid m,\)
则\(\,p\mid(2am^3-n^3)^3\implies p\mid 2am^3-n^3\implies p\mid n.\)
得到矛盾 \(1=\gcd(m,n)\ge p>1\).

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) ...

往前走！若\(a, b, c\ \)是正整数。
试证：\(\sqrt[5]{a+b\sqrt{c}}+\sqrt[5]{a-b\sqrt{c}}=1\ \)无解。

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) ...

往前走！若\(a, b, c\ \)是正整数。
试证：\(\sqrt[5]{a+b\sqrt{c}}+\sqrt[5]{a-b\sqrt{c}}=2\ \)通解公式
LinearRecurrence[{3, -3, 1}, {1, 16, 41}, 16]
{a=1, 16, 41, 76, 121, 176, 241, 316, 401, 496, 601, 716, 841, 976, 1121, 1276}
LinearRecurrence[{3, -3, 1}, {5, 16, 29}, 16]
{b=5, 16, 29, 44, 61, 80, 101, 124, 149, 176, 205, 236, 269, 304, 341, 380}
LinearRecurrence[{3, -3, 1}, {0, 1, 2}, 16]
{c=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

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) ...

回到主帖。通解公式，往下怎么走？

1，\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=2\ \)通解公式
LinearRecurrence[{2, -1}, {1, 4}, 22]
{a=1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64}
LinearRecurrence[{2, -1}, {3, 4}, 22]
{b=3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24}
LinearRecurrence[{2, -1}, {0, 1}, 22]
{c=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21}

2，\(\sqrt[5]{a+b\sqrt{c}}+\sqrt[5]{a-b\sqrt{c}}=2\ \)通解公式
LinearRecurrence[{3, -3, 1}, {1, 16, 41}, 17]
{a=1, 16, 41, 76, 121, 176, 241, 316, 401, 496, 601, 716, 841, 976, 1121, 1276, 1441}
LinearRecurrence[{3, -3, 1}, {5, 16, 29}, 17]
{b=5, 16, 29, 44, 61, 80, 101, 124, 149, 176, 205, 236, 269, 304, 341, 380, 421}
LinearRecurrence[{3, -3, 1}, {0, 1, 2}, 17]
{c=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}

3，\(\sqrt[7]{a+b\sqrt{c}}+\sqrt[7]{a-b\sqrt{c}}=2\ \)通解公式
LinearRecurrence[{4, -6, 4, -1}, {1, 64, 239, 568}, 14]
{a=1, 64, 239, 568, 1093, 1856, 2899, 4264, 5993, 8128, 10711, 13784, 17389, 21568}
LinearRecurrence[{4, -6, 4, -1}, {7, 64, 169, 328}, 14]
{b=7, 64, 169, 328, 547, 832, 1189, 1624, 2143, 2752, 3457, 4264, 5179, 6208}
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 2, 3}, 14]
{c=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}

4，\(\sqrt[9]{a+b\sqrt{c}}+\sqrt[9]{a-b\sqrt{c}}=2\ \)通解公式
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 256, 1393, 4240, 9841}, 12]
{a=1, 256, 1393, 4240, 9841, 19456, 34561, 56848, 88225, 130816, 186961, 259216}
LinearRecurrence[{5, -10, 10, -5, 1}, {9, 256, 985, 2448, 4921}, 12]
{b=9, 256, 985, 2448, 4921, 8704, 14121, 21520, 31273, 43776, 59449, 78736}
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 2, 3, 4}, 12]
{c=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

王守恩

王守恩 发表于 2020-12-16 15:22
回到主帖。通解公式，往下怎么走？

1，\(\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=2\ \)通解公式 ...

举个例子。
\(\sqrt[29]{a+b\sqrt{30}}+\sqrt[29]{a-b\sqrt{30}}=2\ \ a=?\ \ b=?\)