Timeline for Polynomials which always assume perfect power values
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2017 at 23:07 | vote | accept | Stanley Yao Xiao | ||
| Nov 30, 2015 at 8:06 | comment | added | David Zhang | Why the downvote? The question seems reasonable to me. | |
| Nov 30, 2015 at 8:02 | answer | added | user83446 | timeline score: 7 | |
| Nov 30, 2015 at 6:15 | answer | added | nfdc23 | timeline score: 3 | |
| Nov 30, 2015 at 4:04 | comment | added | Stanley Yao Xiao | @VesselinDimitrov Thanks, that argument is perfectly fine for me! | |
| Nov 30, 2015 at 3:56 | comment | added | Vesselin Dimitrov | OK, then there is a $k > 1$ (dividing $d$ as you require) which works for infinitely many $n$ - and you may apply Siegel's theorem (finiteness of integral points of an irrational affine algebraic curve - in this case, the components of $y^k = f(x)$), with the conclusion that $f$ is a $k$-th power. If you want a more "elementary" proof, my best guess is the DLS argument applies just as well in your situation. | |
| Nov 30, 2015 at 3:48 | comment | added | Stanley Yao Xiao | I saw the paper which states your claim in the first paragraph, but that is not what I wanted to know about. I apologize for the vagueness of the original question. | |
| Nov 30, 2015 at 3:47 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
added 229 characters in body
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| Nov 30, 2015 at 3:44 | comment | added | Vesselin Dimitrov | Yes, of course. See Davenport, Lewis, Schinzel, Polynomials of certain special type, Acta Arithmetica IX, 1964 - or Schinzel's collected works. | |
| Nov 30, 2015 at 3:33 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |