Timeline for Parametric solutions of Pell's equation
Current License: CC BY-SA 3.0
7 events
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| Jan 29, 2015 at 18:46 | comment | added | Stefan Kohl♦ | Also, you say that "for generic $a$, the Galois group of $T_n(x)=a$ should be dihedral of order $2n$" -- what I observe from GAP computations is that for $n \geq 5$ this doesn't seem to be true -- e.g. if $T_5(x)-a$ is irreducible, I find that its Galois group is the semidirect product of ${\rm C}_5$ and ${\rm C}_4$, and for $n = 7$ I find it's a semidirect product of ${\rm C}_7$ and ${\rm C}_6$, and for $n=8$ it is 'often' a group of order 32. | |
| Jan 29, 2015 at 0:24 | comment | added | Stefan Kohl♦ | Furthermore, $T_4(x)-a$ is reducible in $\mathbb{Q}[x]$ for 151 among the first 1000 values of $a$, and $T_5(x)-a$ is reducible for 10 among the first 1000 values of $a$, and $T_6(x)-a$ is reducible for 311 among the first 1000 values of $a$, and $T_7(x)-a$ is reducible for 6 among the first 1000 values of $a$, and $T_8(x)-a$ is reducible for 151 among the first 1000 values of $a$. | |
| Jan 29, 2015 at 0:12 | comment | added | Stefan Kohl♦ | I have checked the first 10000 nonsquares in that sequence, and found that for 1793 of them, $T_3(x)-a$ has a rational root. Thus the density seems roughly the same as for the first 60 which you checked. -- Do you still think the density drops to 0? | |
| Jan 28, 2015 at 1:05 | comment | added | David E Speyer | Hmm ... More than I would have naively guessed. I took the list of primitive solutions to Pell's equation for the first $60$ nonsquares oeis.org/A033313 and solved $T_3(x)=a$. There were rational roots in $12$ cases. I still think the density has to drop to $0$, but I wouldn't have guessed it was so common in the first $60$. | |
| Jan 28, 2015 at 0:17 | comment | added | Stefan Kohl♦ | Though I am not sure you are right where you say that for ${\rm deg} P > 2$ there would be "rarely any solutions" -- to me it seems there are plenty of them -- for a list of examples, see here. | |
| Jan 27, 2015 at 21:07 | comment | added | Stefan Kohl♦ | As your "sufficiently divisible" $M$, you can just take $b^2$. Then for ${\rm deg} P = 1$, we obtain the very short expressions $D(x) = b^2x^2+2ax+n$, $P(x) = b^2x+a$, $Q = b$. These expressions yield solutions with much smaller coefficients than Leonardo's. -- Thank you very much for this! I have tabulated the polynomials for $n \leq 100$ here. With our setting $M := b^2$, your solution for ${\rm deg} P = 2$ however gets identical to Leonardo's solution. | |
| Jan 27, 2015 at 19:26 | history | answered | David E Speyer | CC BY-SA 3.0 |