Timeline for Representing $x^3-2$ as a sum of two squares
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 6, 2023 at 9:03 | answer | added | Denis Shatrov | timeline score: 3 | |
| Dec 13, 2022 at 18:40 | comment | added | Bogdan Grechuk | @ Stanley Yao Xiao - do you have any reference to this result? I found paper by H.-E. RICHERT "SELBERG'S SIEVE WITH WEIGHTS" that, under some natural conditions on P, proves that P(n) has a bounded number of prime factors infinitely often. However, how to add the condition that all these factors are congruent to 1 modulo 4? | |
| S Dec 9, 2021 at 10:16 | history | edited | Glorfindel | CC BY-SA 4.0 |
Added polynomials tag
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| S Dec 9, 2021 at 10:16 | history | suggested | Hhhhhhhhhhh |
Added polynomials tag
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| Dec 9, 2021 at 7:51 | review | Suggested edits | |||
| S Dec 9, 2021 at 10:16 | |||||
| Dec 3, 2021 at 8:37 | vote | accept | Bogdan Grechuk | ||
| Dec 2, 2021 at 19:12 | answer | added | Hhhhhhhhhhh | timeline score: 16 | |
| Dec 2, 2021 at 19:05 | comment | added | Stanley Yao Xiao | One should be able to obtain a lower bound which tends to infinity, but not an asymptotic formula, for the number of integers $x$ such that $x^3 - 2$ is a product of at most $r$ primes (for some reasonable $r$, like $r = 11$ I think suffices), each of which is congruent to 1 mod 4. Each such number is then a sum of two squares. | |
| Dec 2, 2021 at 17:39 | comment | added | David E Speyer | As a side note, your $D(t)^3-1$ factors as $a(t) b(t) \overline{a}(t) \overline{b}(t)$ where $a(t) = t^2+2t+1+i$, $b(t) = 4t^4+8t^3+(12+2i)t^2+(8+2i)t+(3+2i)$, and the bar is complex conjugate. If you write $a(t) b(t) = u(t) + i v(t)$, then $u(t)^2+v(t)^2 = D(t)^3-1$. Unfortunately, I have no ideas about the main question. | |
| Dec 2, 2021 at 16:54 | history | asked | Bogdan Grechuk | CC BY-SA 4.0 |