Is this quadratic equality known?
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From: alainverghote on 11 Aug 2010 13:20 Good evening, There are known solutions of form a^2+b^2+c^2+d^2 = f^2 a,b,c,d,f positive integer numbers Is this one of any interest: (x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2+(xy+yz+zx)^2 = (x^2+y^2+z^2)^2 Alain
From: Sherman Forte on 11 Aug 2010 14:06 <alainverghote(a)gmail.com> wrote in message news:d2bfe822-0913-45da-b57d-dfc1bcf5fb1a(a)w30g2000yqw.googlegroups.com... > Good evening, > > There are known solutions of form > a^2+b^2+c^2+d^2 = f^2 a,b,c,d,f positive integer numbers > > Is this one of any interest: > (x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2+(xy+yz+zx)^2 > = > (x^2+y^2+z^2)^2 > > > Alain sure, a = b = c = d = 1 and f = 2
From: alainverghote on 12 Aug 2010 04:17
On 11 août, 20:06, "Sherman Forte" <inva...(a)invalid.com> wrote: > <alainvergh...(a)gmail.com> wrote in message > > news:d2bfe822-0913-45da-b57d-dfc1bcf5fb1a(a)w30g2000yqw.googlegroups.com... > > > Good evening, > > > There are known solutions of form > > a^2+b^2+c^2+d^2 = f^2 a,b,c,d,f positive integer numbers > > > Is this one of any interest: > > (x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2+(xy+yz+zx)^2 > > = > > (x^2+y^2+z^2)^2 > > > Alain > > sure, a = b = c = d = 1 and f = 2 Bonjour Sherman I am pretty sure you might have got out more juice from the equality: (x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2+(xy+yz+zx)^2 = (x^2+y^2+z^2)^2 than you have got. 1) If n is a sum of three squares then n^2 may be written as a sum of four squares. 2)When z^2=2xy (x+y)^4 = (x^2-yz)^2+(y^2-xz)^2+(xy)^2+(xy+yz+zx)^2 ............. Alain |