Other solutions for 80abc(a^2+b^2+c^2) = d^5+e^5?
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From: TPiezas on 29 Nov 2009 11:26 Hello all, There is a basic identity by Lander, (a+b+c)^5 = (-a+b+c)^5 + (a-b+c)^5 + (a+b-c)^5 + 80abc(a^2+b^2+c^2) Problem: Find {a,b,c} such that, 80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1) I found this has an infinite number of non-trivial solns given by, {a,b,c} = {u^5, 25v^5, 50v^5}, and {d,e} = {10u^3v^2, 50uv^4}. for arbitrary {u,v}. Questions: 1) Is there any other parametrizations to eq.1? 2) What other rational {a,b,c;d,e}, excluding {1,25,32; 20,40}, that do not belong to this family? (Enough solns may lead to a pattern.) - Titus
From: alainverghote on 30 Nov 2009 05:20 On 29 nov, 17:26, TPiezas <tpie...(a)gmail.com> wrote: > Hello all, > > There is a basic identity by Lander, > > (a+b+c)^5 = (-a+b+c)^5 + (a-b+c)^5 + (a+b-c)^5 + 80abc(a^2+b^2+c^2) > > Problem: Find {a,b,c} such that, > > 80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1) > > I found this has an infinite number of non-trivial solns given by, > > {a,b,c} = {u^5, 25v^5, 50v^5}, and {d,e} = {10u^3v^2, 50uv^4}. > > for arbitrary {u,v}. Questions: > > 1) Is there any other parametrizations to eq.1? > 2) What other rational {a,b,c;d,e}, excluding {1,25,32; 20,40}, that > do not belong to this family? (Enough solns may lead to a pattern.) > > - Titus Dear Titus, The equality (a+b+c)^5 - (-a+b+c)^5 - (a-b+c)^5 - (a+b-c)^5 = 80abc(a^2+b^2+c^2) is just a particular case of part of form^(2n+1).t^(2p+1).u^(2q+1) of a polynomial g(s,t,u)= (g(s,t,u)-g(-s,t,u)-g(s,-t,u)-g(s,t,-u)+g(-s,-t,u)+g(s,-t,-u)+ g(-s,t,-u) - g(-s,-t,-u))/8 On (s+t+u)^3 we've got: (s+t+u)^3-(-s+t+u)^3-(s-t+u)^3-(s+t-u)^3 = 24stu Alain
From: TPiezas on 30 Nov 2009 22:45 On Nov 30, 4:20 am, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> wrote: > On 29 nov, 17:26, TPiezas <tpie...(a)gmail.com> wrote: > > > > > > > Hello all, > > > There is a basic identity by Lander, > > > (a+b+c)^5 = (-a+b+c)^5 + (a-b+c)^5 + (a+b-c)^5 + 80abc(a^2+b^2+c^2) > > > Problem: Find {a,b,c} such that, > > > 80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1) > > > I found this has an infinite number of non-trivial solns given by, > > > {a,b,c} = {u^5, 25v^5, 50v^5}, and {d,e} = {10u^3v^2, 50uv^4}. > > > for arbitrary {u,v}. Questions: > > > 1) Is there any other parametrizations to eq.1? > > 2) What other rational {a,b,c;d,e}, excluding {1,25,32; 20,40}, that > > do not belong to this family? (Enough solns may lead to a pattern.) > > > - Titus > > Dear Titus, > > The equality > (a+b+c)^5 - (-a+b+c)^5 - (a-b+c)^5 - (a+b-c)^5 = 80abc(a^2+b^2+c^2) > is just a particular case of part of form^(2n+1).t^(2p+1).u^(2q+1) > of a polynomial g(s,t,u)= > (g(s,t,u)-g(-s,t,u)-g(s,-t,u)-g(s,t,-u)+g(-s,-t,u)+g(s,-t,-u)+ > g(-s,t,-u) - g(-s,-t,-u))/8 > On (s+t+u)^3 > we've got: > (s+t+u)^3-(-s+t+u)^3-(s-t+u)^3-(s+t-u)^3 = 24stu > > Alain- Hide quoted text - > > - Show quoted text - I. Yes, the basic form is called Boutin's Identity which generalizes the difference of two squares into a sum and difference of 2^(k-1) kth powers. Thus, (a+b)^2 - (a-b)^2 = 2ab (a+b+c)^3 - (a-b+c)^3 - (a+b-c)^3 + (a-b-c)^3 = 24abc and so on for all kth powers. See Boutin's Identity http://sites.google.com/site/tpiezas/001. II. I found other solns to 80abc(a^2+b^2+c^2) = d^5+e^5. The list {a,b,c}; {d,e} so far is: {1, 25, 32}; {20,40}, {2, 352, 355}; {-328, 388} {4, 125, 155}; {-70, 190} {32, 101, 205}; {-280, 340} and, other than the known family of solns, there doesn't seem to be rhyme or reason to these. - Titus
From: Gerry on 2 Dec 2009 14:48 On Dec 1, 4:45 am, TPiezas <tpie...(a)gmail.com> wrote: > On Nov 30, 4:20 am, "alainvergh...(a)gmail.com" > > > > > > <alainvergh...(a)gmail.com> wrote: > > On 29 nov, 17:26, TPiezas <tpie...(a)gmail.com> wrote: > > > > Hello all, > > > > There is a basic identity by Lander, > > > > (a+b+c)^5 = (-a+b+c)^5 + (a-b+c)^5 + (a+b-c)^5 + 80abc(a^2+b^2+c^2) > > > > Problem: Find {a,b,c} such that, > > > > 80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1) > > > > I found this has an infinite number of non-trivial solns given by, > > > > {a,b,c} = {u^5, 25v^5, 50v^5}, and {d,e} = {10u^3v^2, 50uv^4}. > > > > for arbitrary {u,v}. Questions: > > > > 1) Is there any other parametrizations to eq.1? > > > 2) What other rational {a,b,c;d,e}, excluding {1,25,32; 20,40}, that > > > do not belong to this family? (Enough solns may lead to a pattern.) > > > > - Titus > > > Dear Titus, > > > The equality > > (a+b+c)^5 - (-a+b+c)^5 - (a-b+c)^5 - (a+b-c)^5 = 80abc(a^2+b^2+c^2) > > is just a particular case of part of form^(2n+1).t^(2p+1).u^(2q+1) > > of a polynomial g(s,t,u)= > > (g(s,t,u)-g(-s,t,u)-g(s,-t,u)-g(s,t,-u)+g(-s,-t,u)+g(s,-t,-u)+ > > g(-s,t,-u) - g(-s,-t,-u))/8 > > On (s+t+u)^3 > > we've got: > > (s+t+u)^3-(-s+t+u)^3-(s-t+u)^3-(s+t-u)^3 = 24stu > > > Alain- Hide quoted text - > > > - Show quoted text - > > I. Yes, the basic form is called Boutin's Identity which generalizes > the difference of two squares into a sum and difference of 2^(k-1) kth > powers. Thus, > > (a+b)^2 - (a-b)^2 = 2ab > (a+b+c)^3 - (a-b+c)^3 - (a+b-c)^3 + (a-b-c)^3 = 24abc > > and so on for all kth powers. See Boutin's Identityhttp://sites.google..com/site/tpiezas/001. > > II. I found other solns to 80abc(a^2+b^2+c^2) = d^5+e^5. The list > {a,b,c}; {d,e} so far is: > > {1, 25, 32}; {20,40}, > {2, 352, 355}; {-328, 388} > {4, 125, 155}; {-70, 190} > {32, 101, 205}; {-280, 340} > > and, other than the known family of solns, there doesn't seem to be > rhyme or reason to these. > > - Titus- Hide quoted text - > > - Show quoted text - Hi Titus , two more solutions : {a,b,c; d,e} {1,2,125 ; 10, 50} {a,b,c; d,e} {3,4,240 ; 80,100} As for another equation for : 80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1) i found that : C(a,b,d,e) = c with: P(a,b,d,e)=(9*a^2*b^2*(d^5+e^5)+sqrt(3)*sqrt(a^4*b^4*(25600*a^2*b^2* (a^2+b^2)^3+27*(d^5+e^5)^2)))^(1/3); C(a,b,d,e)=(5^(1/3)*P(a,b,d,e)-40*6^(1/3)*a^2*b^2*(a^2+ b^2)/P (a,b,d,e))/(2*30^(2/3)*a*b); (i don't know how to simplify these) Regards Gerry
From: TPiezas on 3 Dec 2009 22:07
On Dec 2, 1:48 pm, Gerry <gerry...(a)gmail.com> wrote: > On Dec 1, 4:45 am, TPiezas <tpie...(a)gmail.com> wrote: > > > > > > > On Nov 30, 4:20 am, "alainvergh...(a)gmail.com" > > > <alainvergh...(a)gmail.com> wrote: > > > On 29 nov, 17:26, TPiezas <tpie...(a)gmail.com> wrote: > > > > > Hello all, > > > > > There is a basic identity by Lander, > > > > > (a+b+c)^5 = (-a+b+c)^5 + (a-b+c)^5 + (a+b-c)^5 + 80abc(a^2+b^2+c^2) > > > > > Problem: Find {a,b,c} such that, > > > > > 80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1) > > > > > I found this has an infinite number of non-trivial solns given by, > > > > > {a,b,c} = {u^5, 25v^5, 50v^5}, and {d,e} = {10u^3v^2, 50uv^4}. > > > > > for arbitrary {u,v}. Questions: > > > > > 1) Is there any other parametrizations to eq.1? > > > > 2) What other rational {a,b,c;d,e}, excluding {1,25,32; 20,40}, that > > > > do not belong to this family? (Enough solns may lead to a pattern.) > > > > > - Titus > > > > Dear Titus, > > > > The equality > > > (a+b+c)^5 - (-a+b+c)^5 - (a-b+c)^5 - (a+b-c)^5 = 80abc(a^2+b^2+c^2) > > > is just a particular case of part of form^(2n+1).t^(2p+1).u^(2q+1) > > > of a polynomial g(s,t,u)= > > > (g(s,t,u)-g(-s,t,u)-g(s,-t,u)-g(s,t,-u)+g(-s,-t,u)+g(s,-t,-u)+ > > > g(-s,t,-u) - g(-s,-t,-u))/8 > > > On (s+t+u)^3 > > > we've got: > > > (s+t+u)^3-(-s+t+u)^3-(s-t+u)^3-(s+t-u)^3 = 24stu > > > > Alain- Hide quoted text - > > > > - Show quoted text - > > > I. Yes, the basic form is called Boutin's Identity which generalizes > > the difference of two squares into a sum and difference of 2^(k-1) kth > > powers. Thus, > > > (a+b)^2 - (a-b)^2 = 2ab > > (a+b+c)^3 - (a-b+c)^3 - (a+b-c)^3 + (a-b-c)^3 = 24abc > > > and so on for all kth powers. See Boutin's Identityhttp://sites.google.com/site/tpiezas/001. > > > II. I found other solns to 80abc(a^2+b^2+c^2) = d^5+e^5. The list > > {a,b,c}; {d,e} so far is: > > > {1, 25, 32}; {20,40}, > > {2, 352, 355}; {-328, 388} > > {4, 125, 155}; {-70, 190} > > {32, 101, 205}; {-280, 340} > > > and, other than the known family of solns, there doesn't seem to be > > rhyme or reason to these. > > > - Titus- Hide quoted text - > > > - Show quoted text - > > Hi Titus , > > two more solutions : > > {a,b,c; d,e} {1,2,125 ; 10, 50} > {a,b,c; d,e} {3,4,240 ; 80,100} > > As for another equation for : > > 80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1) > > i found that : > > C(a,b,d,e) = c > > with: > > P(a,b,d,e)=(9*a^2*b^2*(d^5+e^5)+sqrt(3)*sqrt(a^4*b^4*(25600*a^2*b^2* > (a^2+b^2)^3+27*(d^5+e^5)^2)))^(1/3); > > C(a,b,d,e)=(5^(1/3)*P(a,b,d,e)-40*6^(1/3)*a^2*b^2*(a^2+ b^2)/P > (a,b,d,e))/(2*30^(2/3)*a*b); > > (i don't know how to simplify these) > > Regards > > Gerry- Hide quoted text - > > - Show quoted text - Thanks for the solns, Gerry. Yes, it can be solved as a cubic in either {a,b,c}. Too bad there's no simple criterion involving the cubic's discriminant D to determine if it has a rational root. (Unlike the quadratic where D simply is made a square.) P.S. I wonder if eq.1, with constraints, can be transformed into an elliptic curve. After all, it has an infinite number of solns. - Titus |