Re: A potential series providing integer solutions to a^2+b^2+c^2...+m^2=n^2

rer wrote:
> For illustrative purposes (and for those of us who are sometimes
> symbollically impaired), would you please give an actual numeric
> example for one of the rows for both 4*F(n+2)*F(n+3)-3 and r(n) =
> (y(n)*y(n+1) - 4)/2 ? Thanks.

First: Do not top post.

Second: Fixed width font with what follows

y(n) = n'th yellow
r(n) = n'th red

n F(n) y(n) r(n)
_________________________________________
1 1 4 10
2 1 6 28
3 2 10 78
4 3 16 206
5 5 26 544
6 8 42 1426
7 13 68 3738
8 21 110 9788

now
1^2+2^2+4^2+6^2+10^2=157=4*F(5)*F(6)-3=y(3)*y(4)-3

and we kow that 2*r(3)+1=157 so
r(3)=(y(3)*y(4)-4)/2 or 78=(10*16-4)/2
similarly
206=(16*26-4)/2
544=(26*42-4)/2
etc.
>
>
> David M Einstein wrote:
> > rer wrote:
> > > I have done some further investigation of a 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
> > > the 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
> >
> > Well, your yellow numbers are just the n+2 nd Fibonacci number where n
> > is the row. That explains the ratio between the yellow terms as the
> > golden ratio. Now the sum of squares of the terms up to and including
> > the yellow term is 4*F(n+2)*F(n+3)-3. So the red term r(n) =
> > (y(n)*y(n+1) - 4)/2. This means that the ratio between successive red
> > terms is the square of the golden ratio, as you predicted.

.

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