Hello Tim, Thank you for the interest. I just moved to Canada and I am still sorting out all my stuff. I will get back to your questions in a few days. :-)
-Tito On Sun, Jun 7, 2009 at 8:59 PM, TimDaly <[email protected]> wrote: > On Jun 4, 12:35 pm, [email protected] wrote: > > Hello all, > > > > Here's a nice identity: > > > > (p+q)^4 + (r-s)^4 = (p-q)^4 + (r+s)^4 > > > > where {p,q,r,s} = {a^7+a^5-2a^3+a, 3a^2, a^6-2a^4+a^2+1, 3a^5} > > > > For similar stuff, you may be interested in "A Collection of Algebraic > > Identities": > > > > http://sites.google.com/site/tpiezas/Home > > > > It's a 200+ page book I wrote and made available there. It starts > > with the basics with 2nd powers and goes up to 8th and higher powers. > > Enjoy. > > > > - Titus > > On the page http://sites.google.com/site/tpiezas/002 > > Theorem: If p^2 + (p+1)^2 = r^2, then q^2 + (q+1)^2 = (p+q+r+1)^2 > where q = 3p+2r+1 > > q:=3p+2r+1 > r:=sqrt(p^2 + (p+1)^2) > q^2 + (q+1)^2 - (p+q+r+1)^2 == (-4r -8p -4)sqrt(2p^2+2p+1)+4r^2+(8p+4) > r > > which is clearly not zero. What am I missing? > > Tim Daly > [email protected] >
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