Only a short while after returning, he has broken the 100k!

Thanks for your answers, and congratulations, Bill!

  • 19
    $\begingroup$ Anyway: $$\rm A\, hearty\, and\, well-earned\, congratulations\, Bill!\ $$ I appreciated so much your answering style: $$\rm \ \ n=3k\!-\!2\!:\ {\rm mod}\,\ \color{#0a0}{d_1:\ a^3\equiv b^2},\, b^3\equiv 1\,\Rightarrow\, a^3b^n\! = \color{#0a0}{a^3} b^{3k-2} \equiv (b^3)^k \equiv 1\,\Rightarrow\, f_n \equiv 0$$ $\endgroup$
    – user93957
    Commented Jan 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. $\endgroup$ Commented Jan 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 Banana
    Commented Jan 9, 2014 at 15:45
  • 2
    $\begingroup$ @GustavoBandeira: OMG, that's gold. :) $\endgroup$
    – Prism
    Commented Jan 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. $\endgroup$ Commented Jan 17, 2014 at 16:39

1 Answer 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
    $\begingroup$ Finally, someone, posted a kind answer. :) $\endgroup$
    – Mikasa
    Commented Jan 16, 2014 at 17:37

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