On Jun 4, 12:35 pm, tpie...@gmail.com 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
d...@axiom-developer.org
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