Veit> Isn't this obsession -- with looking for relationships where there almost certainly aren't any -- a waste of time? [...] rwg> That same day, I sent this to the kids: ---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Thu, Apr 12, 2012 at 6:13 AM Subject: Re: [math-fun] IEG puzzle To: Neil Bickford <techie314@gmail.com>, Julian Ziegler Hunts , Corey Ziegler Hunts Strange: This is two Venusian "days". --Bill

A puzzle I gave my mechanics class:>> The average orbital periods of the Jovian moons Io, Europa and Ganymede > are:>> TI=1.769137786> TE=3.551181041> TG=7.15455296>> These are taken from Wikipedia; the time unit is days.>> 1. (math) On the basis of just these numbers, infer the existence of a > much> longer period (on the order of hundreds of days).

That's a nice puzzle. The first thing I did was to enter TE/TI, TG/TE and TG/TI into a continued fraction calculator, to produce best rational approximations. The first few convergents for TE/TI are: 2/1 275/137 * 3027/1058 3302/1645 And those for TG/TE are: 2/1 137/68 * 3153/1565 3920/1633 Finally, the convergents for TG/TI are: 4/1 89/22 93/23 275/68 * The asterisked convergents suggest a ratio of: TI : TE : TG = (1/275) : (1/137) : (1/68) This means there is a large period of 275 TI = 137 TE = 68 TG. For each of the orbital periods you have provided, this results in the following approximations to the large period: 275 TI = 486.51289115 days 137 TE = 486.511802617 days 68 TG = 486.50960128 days = approx 486.51 days. (Technically, a better method than using continued fractions is the LLL lattice reduction algorithm. However, since the data are so precise, the continued fraction method was adequate.) Sincerely, Adam P. Goucher