Only a short while after returning, he has broken the 100k!
Thanks for your answers, and congratulations, Bill!
$\begingroup$
$\endgroup$
5
-
19$\begingroup$ Anyway: $$\rm A\, hearty\, and\, well-earned\, congratulations\, Bill!\ $$ I appreciated so much your answering style: $$\rm \ \ n=3k\!-\!2\!:\ {\rm mod}\,\ \color{#0a0}{d_1:\ a^3\equiv b^2},\, b^3\equiv 1\,\Rightarrow\, a^3b^n\! = \color{#0a0}{a^3} b^{3k-2} \equiv (b^3)^k \equiv 1\,\Rightarrow\, f_n \equiv 0$$ $\endgroup$– user93957Jan 3, 2014 at 15:19
-
36$\begingroup$ I've deleted the comments about downvotes since such comments tend to become discussions about a users merit that do not really belong on a celebration thread. $\endgroup$– Michael GreineckerJan 4, 2014 at 1:29
-
13$\begingroup$ $$\text{Guess who's back? Back again. Bill is back, tell a friend.}$$ I'm glad he's back. I've even made a silly souvenir. $\endgroup$– Red BananaJan 9, 2014 at 15:45
-
2$\begingroup$ @GustavoBandeira: OMG, that's gold. :) $\endgroup$– PrismJan 13, 2014 at 11:19
-
6$\begingroup$ Thanks! Congratulations and thanks to everyone for their efforts in helping to make MSE the best general-level math forum on the web. $\endgroup$– Bill DubuqueJan 17, 2014 at 16:39
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
After having been away for a year, and upon return, to work this hard on answering questions means there can be one and only one possible answer to this question:
Bill Dubuque is a unique individual who is a truly dedicated mathematics educator.
Thank you for all your hard work.
-
14