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Tue, 09 Sep 2008
Factorials are almost, but not quite, square
$$n! = a^{2} - b^{2}\qquad (*)$$ will always have solutions where b is small compared to a. For example, we have 11! = 6318^{2} - 18^{2}.But to get b=1 might require a lot of luck, perhaps more luck than there is. (Jeremy Kahn once argued that |2^{x} - 3^{y}| = 1 could have no solutions other than the obvious ones, essentially because it would require much more fabulous luck than was available. I sneered at this argument at the time, but I have to admit that there is something to it.) Anyway, back to the subject at hand. Is there an example of n! = a^{2} -1 with n > 7? I haven't checked yet. In related matters, it's rather easy to show that there are no nontrivial examples with b=0. It would be pretty cool to show that equation (*) implied n = O(f(b)) for some function f, but I would not be surprised to find out that there is no such bound. This kept me amused for twenty minutes while I was in line for lunch, anyway. Incidentally, on the lunch line I needed to estimate √11. I described in an earlier article how to do this. Once again it was a good trick, the sort you should keep handy if you are the kind of person who needs to know √11 while standing in line on 33rd Street. Here's the short summary: √11 = √(99/9) = √((100-1)/9) = √((100/9)(1 - 1/100) = (10/3)√(1 - 1/100) ≈ (10/3)(1 - 1/200) = (10/3)(199/200) = 199/60. [ Addendum 20080909: There is a followup article. ] [Other articles in category /math] permanent link |
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