[seqfan] Re: pentolysis: coming apart in five pieces

[franktaw at netscape.net]

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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