[seqfan] Re: pentolysis: coming apart in five pieces
franktaw at netscape.net
franktaw at netscape.net
Sun Apr 18 09:10:51 CEST 2010
Or more simply, all odd squares are congruent to 1 mod 8.
Note that when you subtract 1 from an odd square and divide by 8, you
get an arbitrary triangular number: ((2n+1)^2 - 1) / 8 = n(n+1)/2. So
the question of which numbers congruent to k (mod 8) are the sum of k
[distinct] odd squares is equivalent to the question of whether an
arbitrary number is the sum of k [distinct] triangular numbers. From
A002243, we see that there are only a few numbers that cannot be
represented even as the sum of 3 distinct triangular numbers. There
will be somewhat more that cannot be represent as the sum of 5 such,
because more numbers are too small; but the largest such number won't
be all that large. This means that you only have to check a relatively
few cases to see that any sum of 13 distinct triangular numbers is also
a sum of 5. I haven't actually done such a check, or even determined
how large a value you have to check up to, but it seems fairly likely.
Franklin T. Adams-Watters
-----Original Message-----
From: William Keith <wjk26 at drexel.edu>
> Pentolysis:
> IF a prime number of the form 4n+1 can be written as a sum of
distinct odd
> squares,
> THEN only in a sum of 5 distinct odd squares,
> or only in a sum of 9 distinct odd squares;
> or both in a sum of as well 5 as 13 distinct odd squares,
> but no such prime can be written as a sum of 5 as well as of 9
distinct odd
> squares.
No need for 4n+1 to be prime. This is true for any odd number, since a
sum of 5
distinct odd squares will be 5 or 13 mod 16, while a sum of 9 distinct
odd
squares will be 1 or 9 mod 16.
> And never only as a sum of 13.
Rewriting from 13 to 5, I believe you are always finding that it is the
case
that 9 of the squares add up to one square, yes? Off the top of my
head I'm not
certain this will continue indefinitely, but I notice that if we forget
about
primes or distinctness, 13 = 1+...+1, and 9+1+...+1, and likewise 21 =
9+1+...+1, and 21 = 9+9+1+1+1. Perhaps there is a sort of Pythagorean
9-tuple
in there, modulo 16.
Cordially,
William Keith
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