Timeline for On the equation $a^4+b^4+c^4=2d^4$ in coprime positive integers $a\lt b\lt c$ such that $a+b\ne c$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S Apr 4, 2023 at 1:34 | history | suggested | user178594 | CC BY-SA 4.0 |
Fixed grammar.
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| Apr 3, 2023 at 21:19 | review | Suggested edits | |||
| S Apr 4, 2023 at 1:34 | |||||
| Apr 3, 2023 at 17:25 | comment | added | user178594 | @Tomita What a perfect answer! | |
| Apr 1, 2023 at 17:19 | comment | added | user178594 | I could assume $\gcd(a,b)=1$ in my second previous comment. | |
| Mar 29, 2023 at 18:26 | comment | added | user178594 | Related: mathoverflow.net/questions/440909/…. | |
| Mar 18, 2023 at 11:12 | comment | added | user178594 | Problem: Are almost all integral points to $a^4+b^4+c^4=2d^4$, $0\lt a\lt b\lt c$, of the form $(a,b,a+b,\sqrt{a^2+ab+b^2})$ for some positive integers $a,b$? | |
| Feb 7, 2023 at 4:09 | comment | added | Tomita | If we take the point $Q(m)=\frac{360}{169}$ then we get a small solution $(a,b,c,d)=( 32, 1065, 2321, 1973).$ The large solution corresponds to the point $2Q(m)=\cfrac{-762617488871059540}{36474714629699307}.$ | |
| Feb 6, 2023 at 12:03 | comment | added | Oscar Lanzi | We can actually ask the same question about my answer here. By using a specific form I prove that square roots of integers can give any continued fraction repeat length, but by going for a universally valid structure I sacrifice minimality unless the repeat length happens to be 1 or 3. Might this be a similar trade-off? | |
| Feb 6, 2023 at 10:12 | comment | added | Tito Piezas III | @OscarLanzi: Maybe has to do with the fact it was not brute-force but a rational point on an elliptic curve. Quite similar to this MSE post about co-prime solutions to $a^4+b^4+c^4 = 9d^2$ where an elliptic curve yields $$2682440^4+15365639^4+18796760^4 = 9\times141668657747643^2$$ but the smallest by brute force is only $$155^4+260^4+296^4 = 9\times37747^2.$$ | |
| Jan 15, 2023 at 23:57 | comment | added | Tomita | If everyone were a professional mathematician, maybe $10$ lines answer might be enough. | |
| Jan 15, 2023 at 20:51 | vote | accept | CommunityBot | ||
| Jan 15, 2023 at 13:11 | comment | added | Tomita | I tried to make it understandable without much prerequisite knowledge. | |
| Jan 15, 2023 at 10:55 | comment | added | Oscar Lanzi | Elegant, but why do you suppose this solution is so far from minimal? See the other answer. | |
| Jan 15, 2023 at 0:57 | history | edited | Tomita | CC BY-SA 4.0 |
fixed typo
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| Jan 12, 2023 at 17:10 | vote | accept | CommunityBot | ||
| Jan 15, 2023 at 20:51 | |||||
| Jan 12, 2023 at 10:51 | history | edited | Tomita | CC BY-SA 4.0 |
fixed typo
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| Jan 12, 2023 at 6:58 | history | answered | Tomita | CC BY-SA 4.0 |