A square polynomial equality
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From: alainverghote on 14 May 2010 13:41 Good evening, I wonder if the following identity is known: 4(a^+d^2)(b^2+e^2)(c^2+f^2)+32abcdef = ((a+d)(b+e)(c+f))^2 +((a- d)(b+e)(c- f))^2 +((a+d)(b- e)(c- f))^2 +((a- d)(b- e)(c+f))^2 How can we build such an equality? Best regards, Alain
From: Robert Israel on 14 May 2010 15:30 "alainverghote(a)gmail.com" <alainverghote(a)gmail.com> writes: > I wonder if the following identity is known: > > 4(a^+d^2)(b^2+e^2)(c^2+f^2)+32abcdef > > = > ((a+d)(b+e)(c+f))^2 > +((a- d)(b+e)(c- f))^2 > +((a+d)(b- e)(c- f))^2 > +((a- d)(b- e)(c+f))^2 > How can we build such an equality? Why stop there? sum_e product_{i=1}^n (a_i + e_i b_i)^2 = 2^(n-1) product_{i=1}^n (a_i^2 + b_i^2) + 2^(2n-1) product_{i=1}^n a_i b_i where the sum is over all n-tuples [e_1,...,e_n] of +1 and -1 with product_i e_i = +1. However, since (a_i + e_i b_i)^2 = a_i^2 + b_i^2 + 2 e_i a_i b_i this really comes from sum_e product_{i=1}^n (a_i + e_i b_i) = 2^(n-1) (product_{i=1}^n a_i + product_{i=1}^n b_i) where again the sum is over all n-tuples with product_i e_i = +1. And that's pretty easy to see using symmetry: any term involving, say, a_j and b_k gets multiplied by -1 if you change e_j and e_k to -e_j and -e_k. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: alainverghote on 15 May 2010 11:46
On 14 mai, 21:30, Robert Israel <isr...(a)math.MyUniversitysInitials.ca> wrote: > "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> writes: > > I wonder if the following identity is known: > > > 4(a^+d^2)(b^2+e^2)(c^2+f^2)+32abcdef > > > = > > ((a+d)(b+e)(c+f))^2 > > +((a- d)(b+e)(c- f))^2 > > +((a+d)(b- e)(c- f))^2 > > +((a- d)(b- e)(c+f))^2 > > How can we build such an equality? > > Why stop there? > sum_e product_{i=1}^n (a_i + e_i b_i)^2 > = 2^(n-1) product_{i=1}^n (a_i^2 + b_i^2) + 2^(2n-1) product_{i=1}^n a_i b_i > > where the sum is over all n-tuples [e_1,...,e_n] of +1 and -1 with > product_i e_i = +1. > > However, since (a_i + e_i b_i)^2 = a_i^2 + b_i^2 + 2 e_i a_i b_i > this really comes from > > sum_e product_{i=1}^n (a_i + e_i b_i) > = 2^(n-1) (product_{i=1}^n a_i + product_{i=1}^n b_i) > > where again the sum is over all n-tuples with product_i e_i = +1. > And that's pretty easy to see using symmetry: any term involving, > say, a_j and b_k gets multiplied by -1 if you change e_j and e_k > to -e_j and -e_k. > -- > Robert Israel isr...(a)math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Dear Robert, I started with parity of the product f(s)g(s)h(s) (f(s),g(s),h(s)) <=> (a,b,c) (f(-s),g(-s),h(-s)) <=> (d,e,f) Leading 4(abc+def) =(a+d)(b+e)(c+f) .............. Substitution a=>a^2+d^2, d=>2ad ... yields the given equality. We may use other substitutions: a=>a^3+3a*d^2, d=>3a^2d+d^3 And 4((a^3+3a*d^2)(b^3+3b*e^2)(..)+(d^3+3a^2d)(e^3....)()) = ((a+d)(b+e)(c+f))^3 +((a-d)(b+e)(c-f))^3 +((a+d)(b-e)(c-f))^3 +((a-d)(b-e)(c+f))^3 Best regards, Alain |