Thanks to all for the many useful replies and references, that clarified a lot :). I was playing around with a slightly different aspect of this and a cute little result popped out: If q is a prime power, then (by the freshman's dream) y=\sum (-1)^k x^(q^k) gives a solution for y^q+y=x (the convergence, with respect to the absolute value coming from the `infinte prime' in the PID F_q[t], is for all deg(x)<0 in F_q(t), or even F_q((1/t)) and reasonable extensions thereof.) The method of Eisenstein, as Joerg remarked, would also give z=\sum_{i=0}^\infty \binomial{qi}{i} (-x^(q-1))^i/(1+(q-1)i) as a series solution with integer coefficients, which hence can also be reduced in F_q. It follows that, for instance \binomial{9i,i}*/(1+8i) is 0 mod 3, except when 8i+1=9^k, in which case it is 1 mod 3. Examples of the latter happens when i=1, 10, 91, 820, 7381, etc, which are (of course) precisely the numbers with all ones in their base 9 representation. I am sure there are other ways to prove results like that, but I thought the connection to series solutions was interesting. Best, Ahmad On Thu, Mar 13, 2014 at 11:11 AM, Joerg Arndt <arndt@jjj.de> wrote:
* Ahmad El-Guindy <a.elguindy@gmail.com> [Mar 12. 2014 10:55]:
[...]
So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would
be
defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that?
It looks like http://arxiv.org/abs/1403.2858 is spot on. (Outside my turf, so I cannot rate quality.)
Cheers,
Ahmad _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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