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Topic:
Sum of two squares problem
Replies:
3
Last Post:
Jul 27, 2010 7:44 PM




Re: Sum of two squares problem
Posted:
Jul 27, 2010 9:36 AM


On Jul 27, 2:50 pm, Gerry <ge...@math.mq.edu.au> wrote: > On Jul 27, 9:36 pm, Knud Thomsen <sa...@ktalgorithms.com> wrote: > > > > > For the sum of two squares function (1): > > > r2(c) = number of integer pairs (x,y) for which x^2 + y^2 = c > > > consider the special arguments c' defined by: > > > r2(c') > sup{r2(d): d < c'} > > > A table of c' and r2(c') starts like this: > > >  > > c' r2(c') Factors of r2(c') > >  > > 1 4 1 > > 5 8 5 > > 25 12 5^2 > > 65 16 5 13 > > 325 24 5^2 13 > > 1105 32 5 13 17 > > 4225 36 5^2 13^2 > >  > > <snip!> > > > > > The entries annotated with '?' are tentative, as for the alleged c' > > only r2(c') and r2(c'1) were calculated. The preceding entries, > > however, are based on exhaustive searches. > > > It seems like c' can be factored into primes congruent to 1 modulo 4, > > reminding one of Fermat's theorem on sums of two squares and the > > Brahmagupta?Fibonacci identity. > > > Any comments on the apparent regularity for c'>=5525: each time c' > > doubles, the next prime congruent to 1 modulo 4 is added as a factor? > > > Best regards, > > Knud Thomsen > > Geologist, Denmark > > > References > > 1. Mathworld Sum of Squares Function > > http://mathworld.wolfram.com/SumofSquaresFunction.html > > The column you have labeled Factors of r2(c') is, of course, > actually factorization of c'. On the other hand, when you > write of c' doubling, you actually mean r2(c') doubling. > > Anyway, it is wellknown and not terribly deep that if p > is prime, 1 mod 4, and not a factor of m, then r2(pm) = 2 r2(m). > > Anyway anyway, have a look athttp://www.research.att.com/~njas/sequences/A071383 >  > GM
Many thanks, Gerry, for the references, something for me to dig into. Yes, sorry about the incorrect table header. Too bad I forgot to look up that c' integer sequence myself .. Do you happen to have an (online) reference to a proof of the 'r2(pm) = 2 r2(m)' relation at hand? I mean, it may not be deep, but it isn't obvious to this layman ;).
Knud



