Talk:Casas-Alvero conjecture

Hello fellow Wikipedians,

I have just modified {{plural:1|one external link|1 external links}} on Casas-Alvero conjecture. Please take a moment to review [https://en.wikipedia.org/w/index.php?diff=prev&oldid=749840588 my edit]. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:

• Added archive https://web.archive.org/web/20160304090949/http://www.ems-ph.org/journals/newsletter/pdf/2011-06-80.pdf to http://www.ems-ph.org/journals/newsletter/pdf/2011-06-80.pdf

When you have finished reviewing my changes, please set the checked parameter below to true or failed to let others know (documentation at {{tlx|Sourcecheck}}).

{{sourcecheck|checked=false}}

Cheers.—InternetArchiveBot (Report bug) 12:41, 16 November 2016 (UTC)

If f is a polynomial of degree d, then f^(d-1)(x) is a linear function such that f^(d-1)(a) = 0.

Thus, f^(d-1)(x) = c_d-1*(x - a), where c_n is a constant.

f^(d-2)(x) = c_d-1*(x - a)²/2 = c_d-2*(x - a)²,

and continued integration gives f(x) = c₀*(x - a)^d.

Since it has characteristic zero, a constant cannot be added to each integration since that would negate f^(d-1)(a) = 0.

Note: The only caveat to the conjecture, which it does not explicitly say, is that the function has to be a constant times the power of a linear function.

Am I missing something? BAbdulBaki (talk) 21:39, 3 November 2024 (UTC)

: The conjecture only says some factor is shared, not that that factor must be a constant. The point is to prove that is must be a constant factor. David Malone (talk) 22:34, 3 November 2024 (UTC)

::Thank you! Now it makes sense. Not sure why I didn't read it as that. BAbdulBaki (talk) 00:44, 4 November 2024 (UTC)