Timeline for Small linear relations in unbalanced diophantine equations from primitive Pythagorean triples
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2020 at 9:12 | vote | accept | VS. | ||
| Apr 18, 2020 at 4:30 | answer | added | VS. | timeline score: 0 | |
| Apr 18, 2020 at 3:49 | comment | added | VS. | $(3v_1-v_2)/(2m^2n^2)$ is $(-1,1,a^2-b^2)$. | |
| Apr 17, 2020 at 23:00 | comment | added | VS. | $(3v_1-v_2)/(2m^2n^2)$ gives a shorter basis $(-1,1,(m^4+n^4-6m^2n^2))$. | |
| Apr 17, 2020 at 21:40 | history | edited | VS. | CC BY-SA 4.0 |
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| Apr 17, 2020 at 21:35 | history | edited | VS. | CC BY-SA 4.0 |
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| Apr 17, 2020 at 9:25 | history | edited | VS. | CC BY-SA 4.0 |
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| Apr 17, 2020 at 4:15 | comment | added | VS. | Whatever the reduced basis should be is not clear to me. | |
| Apr 17, 2020 at 4:13 | history | edited | VS. | CC BY-SA 4.0 |
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| Apr 17, 2020 at 4:00 | history | edited | VS. | CC BY-SA 4.0 |
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| Apr 17, 2020 at 3:54 | history | edited | VS. | CC BY-SA 4.0 |
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| Apr 17, 2020 at 2:38 | history | edited | VS. | CC BY-SA 4.0 |
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| Apr 17, 2020 at 2:32 | history | asked | VS. | CC BY-SA 4.0 |