Pythagorean Quads, Quints, ...

[rer]

Pythagorean triples provide integer solutions to the equation:

a^2+b^2=c^2

Consider the equations:

a^2+b^2+c^2...+m^2=n^2

Propositions:

1) all such equations have nonrepetitive (i.e., a<>b<>c...<>m) multiple
solutions such that a,b,c...m,n are integer (please note the series
shown at the bottom of this page,
http://www.maplenet.net/~reriker/madpythag.html -- beginning with b=2,
c=4, where the term m-1 is formed from the sum of the term m-3 + the
term m-2 ; e.g., 2+4=6, 4+6=10, 6+10=16, 10+16=26, and 16+26=42 ; the
next term in the series should be 26+42=68 ; and, indeed,
1^2+2^2+4^2+6^2+10^2+16^2+26^2+42^2+68^2+3738^2=3739^2)

2) all squares of odd integers greater than 3 may be formed by the sum
of nonrepetitive integer squares

3) more speculative, all squares of odd integers greater than 3 may be
formed by 5 or less nonrepetitive integer squares

Some examples shown here:
http://www.maplenet.net/~reriker/madpythag.html

[Gerry Myerson]

In article <1124172969.8...@g49g2000cwa.googlegroups.com>,
"rer" <rer...@maplenet.net> wrote:

> 3) more speculative, all squares of odd integers greater than 3 may be
> formed by 5 or less nonrepetitive integer squares

Gordon Pall, On sums of squares, American Mathematical Monthly 40
(1933) 10-18, Theorem 2 (p. 13):

The only integers n > 0 not sums of four unequal squares are
4^h a, where h = 0, 1, 2, ..., and
a = 1, 3, 5, 7, 9, 11, 13, 15, 19, 23, 25, 27, 31, 33, 37, 43,
47, 55, 67, 73, 97, 103, 2, 6, 10, 18, 22, 34, 58, 82.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

[Minus XVII]

wait; the only integers not the sum of three second-powers (or
"skwares"), are those of the form, 4^h + 7, if
I properly recall (see _The Theory of Numbers_
by Hardy and Wright e.g.); that's the four-skwares problem.

there is an infinite family of sums of second powers
of n consecutive integers,
adding to a second power, each of which has an infinity
of solutions (like in the pythagorean triples);
I forget the reference.

--Hemp for Haemarrhoids (Bogart that poultice) !!
http://members.tripod.com/~american_almanac

[Gerry Myerson]

In article <1124239171.0...@g47g2000cwa.googlegroups.com>,
"Minus XVII" <Qnc...@netscape.net> wrote:

> wait; the only integers not the sum of three second-powers (or
> "skwares"), are those of the form, 4^h + 7, if
> I properly recall (see _The Theory of Numbers_
> by Hardy and Wright e.g.);

Not quite.

> that's the four-skwares problem.

Not quite.

The numbers not the sum of 3 squares are 4^h x (8 k + 7).

The four squares theorem says even these numbers are sums
of four squares.

But the OP was asking about sums of distinct squares,
which is a bit harder.

[Minus XVII]

ah, so; the first such number is 7 (2^2 + 1^2 + 1^2 + 1^2); anyway,
second-powers of consecutive integers fit the bill.

[rer]

I have done some further investigation of the series that appears to
satisfy the requirements of providing integer solutions to the equation
a^2+b^2+c^2...+m^2=n^2

It appears that the ratios of the terms m of this series come close to
1 + the Golden Ratio (e.g., 2.61804), which is useful in predicting the
next term m in the series.

I have updated the original with a table to better illustrate this:

http://www.maplenet.net/~reriker/madpythag.html

I am going to repost to the larger list, so my apologies if you see
this twice.

[Minus XVII]

extraordinary!

phi +1 equals phi^2, showing that
2.618... is as fundamental as 1.618...;
may even have found a fast convergence on the ratio !?

--Hemp for Haemarrhoids (Bpgart that Poultice, Friend) !!
http://members.tripod.com/~american_almanac

[rer]

I have found a second series:

2 4 5 22 23
2 4 5 10 72 73
2 4 5 10 14 170 171
2 4 5 10 14 24 458 459
2 4 5 10 14 24 38 1180 1181
etc.

where term m (((m*((m+(term m-1))+4)/2 (e.g., for row 4
(24*(24+14))+4)/2=458)

Although it would seem a bit preliminary, my suspicion is, like
Pythagorean triples, there are an endless number of these. In these
first two, the difference between n-m = 1. I am going to investigate
to see if I can find a series where n-m=2 (e.g., similar to the
Pythagorean triple 8,15,17). If I can find one, it will be interesting
to see what the ratio is there.

[rer]

As I suspected, it appears very easy to form such series.

Consider a^2+b^2+c^2+d^2=e^2,

let a = an odd integer greater than 1
let b = a+1
let c = 2a

Then d = ((c*(b+c))/2
and e = d+1

Then, consider a^2+b^2+c^2+d^2+e^2=f^2

a, b, and c are as above
now let d = b+c (the beginning of an off-set Fibonacci series)
then e = (d*(c+d))/2
and f = e+1

Then, consider a^2+b^2+c^2+d^2+e^2+f^2=g^2

a, b, c, and d are as above
now let e = c+d (the second member of the off-set Fibonnacci series)
then f = (e*(d+e))/2
and g = f+1

etc.

Some examples are here:

http://www.maplenet.net/~rerik er/madpythag2.html

[rer]

An outgrowth of exploring this is another observation (perhaps trivial,
preceded, or just plain wrong) about the structure of the primitive
versions of these Pythagorean progressions:

http://www.maplenet.net/~reriker/madpythag3.html

[rer]

I was able to find a couple of series where n-m=2 fairly easily.
However, the ratio did not appear to be particularly interesting -- it
looks like it just approaches 1. Two examples may be found here
(tables 3 & 4 -- http://www.maplenet.net/~reriker/madpythag.html ) and
the beginning part of table 3 is:

a b c d e f g h
row1 2 6 8 25 27
row2 2 6 8 12 61 63
row3 2 6 8 12 16 125 127
row4 2 6 8 12 16 20 225 227

The m term (e.g., 61, 125, 225) can be determined by the formula:

for row2, e2=d1+((d2/2)^2)

e.g.
61 = 25 + ((12/2)^2)
61 = 25 + 36

for row3, f3=e2+((e3/2)^2)

125 = 61 + ((16/2)^2)
125 = 61 + 64

etc.

[rer...@gmail.com]

I have moved these pages to:

http://mysite.verizon.net/reriker/