Timeline for Parametric solutions of Pell's equation
Current License: CC BY-SA 3.0
34 events
| when toggle format | what | by | license | comment | |
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| Jan 27 at 14:08 | history | protected | CommunityBot | ||
| Feb 9, 2017 at 0:11 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Updated a link.
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| S May 2, 2016 at 4:23 | history | suggested | user57432 | CC BY-SA 3.0 |
link edited
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| May 2, 2016 at 4:14 | review | Suggested edits | |||
| S May 2, 2016 at 4:23 | |||||
| Feb 17, 2015 at 23:38 | comment | added | Stefan Kohl♦ | @Leonardo: Very interesting! – I have now added a reference to your note to your first answer, and accepted that answer. | |
| Feb 17, 2015 at 23:36 | vote | accept | Stefan Kohl♦ | ||
| Feb 17, 2015 at 23:35 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Update: added a link to Leonardo Zapponi's note answering the question.
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| Feb 16, 2015 at 20:37 | comment | added | Leonardo | Actually, as shown in the remark at the end of the (updated) note, if we don't assume that $n$ is square-free then any degree can occur. | |
| Feb 16, 2015 at 19:38 | comment | added | Leonardo | I actually believe that in the general situation the degree of $P$ is unbounded. Take for example $n=5043$ and the fundamental solution $a=5042$ and $b=71$. We then have the parametric solution $D=2825761X^2+275684X+5043$, $P=12464273528384X^7+4256093399936X^6+609867042064X^5+47472784800X^4+2165222136X^3+57799504X^2+835457X+5042$ and $Q=7414796864X^6+2170184448X^5+259142960X^4+16137600X^3+552024X^2+9824X+71$. | |
| Feb 16, 2015 at 18:23 | comment | added | Leonardo | This is not yet clear... I will work on the question. | |
| Feb 16, 2015 at 18:18 | comment | added | Stefan Kohl♦ | @Leonardo: I see. -- And which degrees can occur for $n$ which are not square-free? | |
| Feb 16, 2015 at 16:24 | comment | added | Leonardo | Dear Stefan, the possible degrees are $1,2,3$ and $6$ if $n$ is square-free. The examples you give in the above comment correspond respectively to $n=12$ (for $x=0$) and $n=500$ (for $x=-1$), which are not square-free. | |
| Feb 16, 2015 at 14:48 | comment | added | Stefan Kohl♦ | @Leonardo: Thank you! But hmm: the degrees $4$ and $5$ do occur as well -- for example, solutions are $D = 16x^2+16x+12$, $P = 32x^4+64x^3+64x^2+32x+7$, $Q = 8x^3+12x^2+8x+2$ and $D = 625x^2+125x$, $P = 1600000x^5+800000x^4+140000x^3+10000x^2+250x+1$, $Q = 64000x^4+25600x^3+3360x^2+160x+2$. -- How do you explain these? | |
| Feb 16, 2015 at 13:57 | comment | added | Leonardo | The note is now available here. It is a draft version, which might soon be modified. It turns out that the possible degrees for $P$ are $1,2,3$ and $6$ (and they actually all occur). | |
| Feb 14, 2015 at 23:23 | comment | added | Leonardo | Dear Stefan, I know have a complete proof that if $n$ is square-free then the degree of $P$ is bounded by $6$. I'm writing a short, summarizing note; you will receive it shortly. | |
| Feb 14, 2015 at 21:50 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Added a short notice.
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| Feb 3, 2015 at 12:07 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Added a note on what remains to be done in order to complete Leonardo's answers to the question.
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| Feb 2, 2015 at 14:51 | answer | added | Leonardo | timeline score: 4 | |
| Jan 27, 2015 at 21:10 | comment | added | user157310 | I think i have done these equations with D=constant; In this special case then P and Q can be of any degree. But one can get an infinite number of solutions with the degree being always equal to 2. I am sure this is not what you asked but.......the simplest answer can always be expressed with a polynomial of degree 2.{Note here that D has degree zero!!!] | |
| Jan 27, 2015 at 19:26 | answer | added | David E Speyer | timeline score: 3 | |
| Jan 27, 2015 at 12:20 | answer | added | Leonardo | timeline score: 19 | |
| Jan 27, 2015 at 7:21 | answer | added | individ | timeline score: 1 | |
| Jan 27, 2015 at 5:15 | answer | added | Will Jagy | timeline score: 4 | |
| Jan 27, 2015 at 0:26 | comment | added | P Vanchinathan | @StefanKohl: I was referring to rational parametrization obtained by projection for any conic; the correspondence, barring a few finite exceptions, between rational points on $P^1$ and the curve will in general be given by polynomials with rational coefficients, I am not sure now it is possible to guarantee integer coefficients. | |
| Jan 26, 2015 at 23:30 | comment | added | Dror Speiser | @StefanKohl: Oh! Right. I missed that. Maybe it is bounded! | |
| Jan 26, 2015 at 23:28 | comment | added | Stefan Kohl♦ | @DrorSpeiser: A main point is that you don't have $x$'es at every coefficient -- many (most) coefficients are constant throughout a series, and only some are variable and contribute to the degree. | |
| Jan 26, 2015 at 23:27 | comment | added | Dror Speiser | Also, if the answer is yes, then the degress are definitely unbounded, since if $\sqrt{n}$ has a long period for its continued fraction, then $\sqrt{D}$ also has a long period, so $P,Q$ will be of large degree. | |
| Jan 26, 2015 at 23:21 | comment | added | Dror Speiser | This is a guess, but I think the following is true: say $\sqrt{n}=[a_0;\overline{a_1,a_2,...,a_k}]$ - the periodic contiued fraction expansion. Then let $D(x)=[a_0x;\overline{a_1x,a_2x,...,a_kx}]^2$. I think $D(x)$ is integral, and from the convergents you get $P$ and $Q$. Maybe this $D$ doesn't work, but I definitely think something like this is the way to go. | |
| Jan 26, 2015 at 18:07 | comment | added | GH from MO | Well, saying that "$n=13$ belong to two series" suggests that you fix $n$ and look for parametric solutions of $a^2-nb^2=1$ in the remaining variables $a$ and $b$. Perhaps saying that the "initial value $n=13$ belongs to two series" would be more appropriate. This just a language thing, of course, because your question is clear about what you are after. | |
| Jan 26, 2015 at 17:32 | comment | added | Stefan Kohl♦ | @GHfromMO: The latter is true (of course!) -- but what does it have to do with the former? -- In what way do you think it is "misleading" to say that $n=13$ belongs to two series? | |
| Jan 26, 2015 at 17:29 | comment | added | GH from MO | I think it is misleading to say that $n=13$ belongs to two series. Changing the parameter $k$ in your example changes all three variables in the Pell equation $a^2-nb^2=1$, including $n$. | |
| Jan 26, 2015 at 16:24 | comment | added | Stefan Kohl♦ | @PVanchinathan: Sorry, but I don't understand your comment -- could you elaborate? | |
| Jan 26, 2015 at 14:31 | comment | added | P Vanchinathan | This is curve defined over Q and of degree 2. So it will be a rational curve if it has a point defined over Q. For the so called Pell's equation the hypothesis holds. | |
| Jan 26, 2015 at 12:52 | history | asked | Stefan Kohl♦ | CC BY-SA 3.0 |