Timeline for Pythagorean 5-tuples
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Dec 18, 2017 at 11:42 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typo corrected
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| S Nov 1, 2017 at 3:22 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Added some paragraph breaks (to already-bumped question)
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| Nov 1, 2017 at 1:55 | review | Suggested edits | |||
| S Nov 1, 2017 at 3:22 | |||||
| May 14, 2011 at 13:48 | comment | added | Geoff Robinson |
Yes, that is true. But I think the fact that $\sigma$ is an outer automorphism shows that a product like $p(u).p(u)\sigma$ can't usually be expressed in the form $p(u)v^{-1}p(u)v$ for some unit of $\mathbb{H}$.
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| May 14, 2011 at 8:45 | comment | added | mikhail skopenkov | To Remark of May, 13: probably we do not even need to assume that $\sigma:H(u)\to H(u)$ is an automorphism - norm conservation is enough, e.g., $\sigma:x+iy+jz+kt\mapsto x-iy+jz+kt$. | |
| May 13, 2011 at 22:06 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Added remark about new type of solution.
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| May 7, 2011 at 21:06 | vote | accept | mikhail skopenkov | ||
| May 7, 2011 at 21:06 | history | bounty ended | mikhail skopenkov | ||
| May 7, 2011 at 13:35 | comment | added | mikhail skopenkov | Answer of doetoe can be a candidate for a counter-example to 2). | |
| May 6, 2011 at 13:41 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typos
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| May 6, 2011 at 13:34 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Corrected latex
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| May 3, 2011 at 7:28 | comment | added | mikhail skopenkov | Extended answer is beautiful! It provides a large set S of solutions, and now we know that S is a sub-semigroup of the semigroup of all the solutions. Are there any ideas if 1) there are examples of Pythagorean 5-tuples not in S; 2) we need arbitrary n, or any element of S can be represented as, say, (x,y,z,t)=p1p2p2p1, w=|p1p2|^2, where p1,p2 belong to H[u]? | |
| May 3, 2011 at 6:44 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
minor typo
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| May 3, 2011 at 2:37 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Minor textual changes
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| May 2, 2011 at 22:58 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Minor textual changes
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| Apr 25, 2011 at 10:22 | comment | added | mikhail skopenkov | @aaron The formula for polynomials solutions of $x^2+y^2=z^2$ is $x=d(p^2-q^2)$, $y=2dpq$, $z=d(p^2+q^2)$, where p,q,d are arbitrary polynomials. It is known that a similar formula for n=1 does not provide all 4-tuples [4, Theorem 2.2 and end of Section 3.2]. There are reasons to believe that there are also 5-tuples not obtained in this way but I cannot give an example immediately. | |
| Apr 24, 2011 at 19:17 | comment | added | Aaron Meyerowitz | @mikhail What is a polynomial 5-tuple not obtained from this and what is the formula for polynomial solutions to $x^2+y^2=z^2$ that you have in mind? | |
| Apr 24, 2011 at 11:54 | comment | added | mikhail skopenkov | Thank you. Both your approach and the formula from Mordell's "Diophantine Equations" (chapter 3) describes the solutions obtained by "inverse stereographic projection" of the 3-space with the coordinates $p:q:r:s$ to the 3-sphere $x^2+y^2+z^2+t^2=w^2$: $(p^2-q^2-r^2-s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$. This provides a "large" set of Pythagorean 5-tuples of polynomials, if one takes arbitrary polynomials p,q,r,s. Not every Pythagorean 5-tuple of polynomials can be obtained in this way, thus other formulas are also of interest. | |
| Apr 24, 2011 at 10:54 | history | answered | Geoff Robinson | CC BY-SA 3.0 |