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Apr 17, 2017 at 14:23 history reopened Jeremy Rickard
Franz Lemmermeyer
Michael Albanese
Pace Nielsen
Peter Mueller
Apr 17, 2017 at 13:58 comment added Pace Nielsen @trenta3 Thank you for clarifying your meaning. I'm voting to reopen.
Apr 17, 2017 at 10:14 review Reopen votes
Apr 17, 2017 at 14:23
Apr 17, 2017 at 10:03 comment added Felipe Voloch The answers given explain why there are only finitely many solutions for any fixed $k>2$. But, the clarified question asks about varying $k$. I don't recall seeing a general result like that, so it may be open. The finiteness follows from the abc conjecture applied to the displayed equation in (the current version of) Joe's answer.
Apr 17, 2017 at 9:53 comment added trenta3 @PaceNielsen I've rephrased the question. $y$ was not meant to be a polynomial and I was asking about integer solutions.
Apr 17, 2017 at 9:50 history edited trenta3 CC BY-SA 3.0
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Apr 17, 2017 at 9:14 history closed Pace Nielsen
Franz Lemmermeyer
Peter Mueller
Jan-Christoph Schlage-Puchta
Stefan Kohl
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Apr 17, 2017 at 7:47 comment added Peter Mueller I and others misread the question, apparently automatically converting it to a proper question. As such it of course should be closed. Maybe not deleted, for then Joe Silverman's detailed answer would be gone.
Apr 17, 2017 at 6:41 comment added Franz Lemmermeyer The 4 upvotes must be for perfect formatting, right?
Apr 17, 2017 at 2:07 review Close votes
Apr 17, 2017 at 9:14
Apr 17, 2017 at 1:49 comment added Pace Nielsen It is impossible for $f(x)$ to both be irreducible, and a product of two or more nontrivial polynomials.
Apr 16, 2017 at 22:59 answer added Joe Silverman timeline score: 14
Apr 16, 2017 at 22:48 answer added Peter Mueller timeline score: 6
Apr 16, 2017 at 22:41 review First posts
Apr 16, 2017 at 22:46
Apr 16, 2017 at 22:36 history asked trenta3 CC BY-SA 3.0