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More $a^3+b^3+c^3=(c+1)^3,$ and $\sqrt[3]{\cos\tfrac{2\pi}7}+\sqrt[3]{\cos\tfrac{4\pi}7}+\sqrt[3]{\cos\tfrac{8\pi}7}=\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$

score 214 · Answer 1 · 2026-03-28 09:57:36

214

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More $a^3+b^3+c^3=(c+1)^3,$ and $\sqrt[3]{\cos\tfrac{2\pi}7}+\sqrt[3]{\cos\tfrac{4\pi}7}+\sqrt[3]{\cos\tfrac{8\pi}7}=\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$

Published on 28 Mar 2026 - 9:57

score 241 · Answer 2 · 2026-04-02 17:34:28

241

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Mordell's Theorem and Fermat's Last Theorem for $n=4$: A general method?

Published on 02 Apr 2026 - 17:34

score 181 · Answer 3 · 2023-11-10 19:30:31

181

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$(a-1)(b-1)(c-1)=abc-101$ how to find all integer solutions for this one

Published on 10 Nov 2023 - 19:30

score 82 · Answer 4 · 2023-11-10 20:28:51

82

Views

What remains true from Wiles' Theorem for complex integers?

Published on 10 Nov 2023 - 20:28

score 39 · Answer 5 · 2023-11-13 17:21:25

39

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Can relative (or even absolute) quotient size be calculated from a list of polynomials which are multiples of a given variable?

Published on 13 Nov 2023 - 17:21

score 127 · Answer 6 · 2023-11-19 00:57:46

127

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Integral Solutions of $x^3+y^3+z^3+u^3=25$

Published on 19 Nov 2023 - 0:57

score 48 · Answer 7 · 2026-04-03 09:24:41

48

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Diophantus's Problem V-10

Published on 03 Apr 2026 - 9:24

score 308 · Answer 8 · 2023-11-20 12:51:50

308

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Sum of integers=sum of their reciprocals

Published on 20 Nov 2023 - 12:51

score 168 · Answer 9 · 2023-11-22 14:33:18

168

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Solving this three variable diophantine equation.

Published on 22 Nov 2023 - 14:33

score 102 · Answer 10 · 2023-11-24 05:03:22

102

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Finding the number of positive integer solutions to $abc = 16(a+b+c)$

Published on 24 Nov 2023 - 5:03

score 557 · Answer 11 · 2023-11-25 16:29:58

557

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Sum of integers =sum of their reciprocals special case

Published on 25 Nov 2023 - 16:29

score 168 · Answer 12 · 2023-11-26 14:51:01

168

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when is $(5a^2-3)(a^2+1)$ a perfect square

Published on 26 Nov 2023 - 14:51

score 44 · Answer 13 · 2023-11-26 22:44:02

44

Views

Help with diophantine equation: $m^2 -2n^2 = 10$

Published on 26 Nov 2023 - 22:44

score 225 · Answer 14 · 2023-11-27 10:37:00

225

Views

Extended Fermat's Last Theorem

Published on 27 Nov 2023 - 10:37

score 194 · Answer 15 · 2026-03-27 06:09:16

194

Views

Is there any odd $n\in \mathbb{N}$ satisfying $n=e^2-f^2=g^2-h^2=k^2-l^2$ and $e^2f^2=g^2h^2+k^2l^2$ where $e, f, g, h, k, l~(>1)\in \mathbb{N}$?

Published on 27 Mar 2026 - 6:09

List Question

More $a^3+b^3+c^3=(c+1)^3,$ and $\sqrt[3]{\cos\tfrac{2\pi}7}+\sqrt[3]{\cos\tfrac{4\pi}7}+\sqrt[3]{\cos\tfrac{8\pi}7}=\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$

Mordell's Theorem and Fermat's Last Theorem for $n=4$: A general method?

$(a-1)(b-1)(c-1)=abc-101$ how to find all integer solutions for this one

What remains true from Wiles' Theorem for complex integers?

Can relative (or even absolute) quotient size be calculated from a list of polynomials which are multiples of a given variable?

Integral Solutions of $x^3+y^3+z^3+u^3=25$

Diophantus's Problem V-10

Sum of integers=sum of their reciprocals

Solving this three variable diophantine equation.

Finding the number of positive integer solutions to $abc = 16(a+b+c)$

Sum of integers =sum of their reciprocals special case

when is $(5a^2-3)(a^2+1)$ a perfect square

Help with diophantine equation: $m^2 -2n^2 = 10$

Extended Fermat's Last Theorem

Is there any odd $n\in \mathbb{N}$ satisfying $n=e^2-f^2=g^2-h^2=k^2-l^2$ and $e^2f^2=g^2h^2+k^2l^2$ where $e, f, g, h, k, l~(>1)\in \mathbb{N}$?

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