Re: (c+y)^2 + (-y+c-x)^2 + (x+c)^2 = (c-y)^2 + (y+c+x)^2 + (-x+c)^2
From: RASINGER (helmut_rasinger_at_hotmail.com)
Date: 06/10/04
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- In reply to: William Elliot: "Re: (c+y)² + (-y+c-x)² + (x+c)Q_=3D_=28c-y=29=B2_+_=28y+c+x=29=B2_+_=28-x+c?= Q=B2?="
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Date: Thu, 10 Jun 2004 23:30:21 +0200
Sorry, mea culpa
I didnt know about charset problem in newsgroups
"William Elliot" <marsh@privacy.net> a écrit dans le message de
news:20040609231020.G22357@agora.rdrop.com...
> On Thu, 10 Jun 2004, Gerry Myerson wrote:
> > "RASINGER" <helmut_rasinger@hotmail.com> wrote:
> >
> > > (c+y)² + (-y+c-x)² + (x+c)² = (c-y)² + (y+c+x)² + (-x+c)²
> >
> > I take it that what looks like quotation marks on my screen
> > are really supposed to be superscript twos, so we're talking
> > about a sum-of-squares identity. See Gloden, Mehrgradige Gleichungen.
> >
> On my screen those "quotation marks" appeared as shaded rectangles.
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