A constrained simple equality
in [Math]
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From: alainverghote on 22 May 2010 12:58 On 22 mai, 10:27, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Fri, 21 May 2010, alainvergh...(a)gmail.com wrote: > > Form x^2 + 2y^2 = z^2 + 2t^2 , {x,t} given > > > In case we wish four distinct squares, example for {2,3} > > 2^2 + 2*13^2 = 18^2 + 2*3^2 > > we need another solution, > > a simple one > > could be {x,y,z,t} = {x,3t+2x,4t+3x,t} > > y = 2x + 3t; z = 3x + 4t is a solution to x^2 + 2y^2 = z^2 + 2t^2? > > x^2 + 2(4x^2 + 12xt + 9t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 > x^2 + 8x^2 + 24xt + 18t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 > > Ok, that is correct. So is y = t, z = x a solution. > > You are looking for elements r,s in the ring R = Z[sqr 2] > for which norm(r) = n(r) = n(s). > n(r) = rr* where (n + m.sqr 2)* = n - m.sqr 2 for n,m in Z. > > There should be an infinite sequence of such numbers. > Find some u /= +-1 in R with n(u) = 1. > For example 3 + 2.sqr 2 or 3 - 2.sqr 2. > Then for all j in N, n(r) = n(u^j r). Bonsoir William, For the given example x^2 + 2y^2 = z^2 + 2t^2 {x,t} = {2,3} which other solutions do you propose? Alain
From: William Elliot on 23 May 2010 00:21 On Sat, 22 May 2010, alainverghote(a)gmail.com wrote: > On 22 mai, 10:27, William Elliot <ma...(a)rdrop.remove.com> wrote: >> On Fri, 21 May 2010, alainvergh...(a)gmail.com wrote: >>> Form �x^2 + 2y^2 = z^2 + 2t^2 , {x,t} given >> >>> In case we wish four distinct squares, example for {2,3} >>> �2^2 + 2*13^2 = 18^2 + 2*3^2 >>> we need another solution, >>> a simple one >>> could be {x,y,z,t} = {x,3t+2x,4t+3x,t} >> >> y = 2x + 3t; �z = 3x + 4t is a solution to x^2 + 2y^2 = z^2 + 2t^2? >> >> x^2 + 2(4x^2 + 12xt + 9t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 >> x^2 + 8x^2 + 24xt + 18t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 >> >> Ok, that is correct. �So is y = t, z = x a solution. >> >> You are looking for elements r,s in the ring R = Z[sqr 2] >> � � � � for which norm(r) = n(r) = n(s). >> n(r) = rr* where (n + m.sqr 2)* = n - m.sqr 2 for n,m in Z. >> >> There should be an infinite sequence of such numbers. >> Find some u /= +-1 in R with n(u) = 1. >> For example 3 + 2.sqr 2 or 3 - 2.sqr 2. >> Then for all j in N, n(r) = n(u^j r). > > For the given example > x^2 + 2y^2 = z^2 + 2t^2 {x,t} = {2,3} > > which other solutions do you propose? > For each n in N, calculate r_n = (2 + 3.sqr 2)(3 +- 2.sqr)^n r_n = a_n + b_n.sqr 2 for some a_n, b_n in Z. Set y = a_n, z = b_n.
From: alainverghote on 23 May 2010 07:00 On 23 mai, 06:21, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Sat, 22 May 2010, alainvergh...(a)gmail.com wrote: > > On 22 mai, 10:27, William Elliot <ma...(a)rdrop.remove.com> wrote: > >> On Fri, 21 May 2010, alainvergh...(a)gmail.com wrote: > >>> Form x^2 + 2y^2 = z^2 + 2t^2 , {x,t} given > > >>> In case we wish four distinct squares, example for {2,3} > >>> 2^2 + 2*13^2 = 18^2 + 2*3^2 > >>> we need another solution, > >>> a simple one > >>> could be {x,y,z,t} = {x,3t+2x,4t+3x,t} > > >> y = 2x + 3t; z = 3x + 4t is a solution to x^2 + 2y^2 = z^2 + 2t^2? > > >> x^2 + 2(4x^2 + 12xt + 9t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 > >> x^2 + 8x^2 + 24xt + 18t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 > > >> Ok, that is correct. So is y = t, z = x a solution. > > >> You are looking for elements r,s in the ring R = Z[sqr 2] > >> for which norm(r) = n(r) = n(s). > >> n(r) = rr* where (n + m.sqr 2)* = n - m.sqr 2 for n,m in Z. > > >> There should be an infinite sequence of such numbers. > >> Find some u /= +-1 in R with n(u) = 1. > >> For example 3 + 2.sqr 2 or 3 - 2.sqr 2. > >> Then for all j in N, n(r) = n(u^j r). > > > For the given example > > x^2 + 2y^2 = z^2 + 2t^2 {x,t} = {2,3} > > > which other solutions do you propose? > > For each n in N, calculate > > r_n = (2 + 3.sqr 2)(3 +- 2.sqr)^n > > r_n = a_n + b_n.sqr 2 for some a_n, b_n in Z. > > Set y = a_n, z = b_n.- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Thanks for your answering me. My particular solution was based on the following identity: (a+2b+2c)^2+(2a+b+2c)^2+c^2 = a^2 +b^2 +(2a+2b+3c)^2 for a=b c^2 +2(3a+2c)^2 = (4a+3c)^2+2a^2 Alain
From: alainverghote on 25 May 2010 10:12
On 23 mai, 13:00, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> wrote: > On 23 mai, 06:21, William Elliot <ma...(a)rdrop.remove.com> wrote: > > > > > > > On Sat, 22 May 2010, alainvergh...(a)gmail.com wrote: > > > On 22 mai, 10:27, William Elliot <ma...(a)rdrop.remove.com> wrote: > > >> On Fri, 21 May 2010, alainvergh...(a)gmail.com wrote: > > >>> Form x^2 + 2y^2 = z^2 + 2t^2 , {x,t} given > > > >>> In case we wish four distinct squares, example for {2,3} > > >>> 2^2 + 2*13^2 = 18^2 + 2*3^2 > > >>> we need another solution, > > >>> a simple one > > >>> could be {x,y,z,t} = {x,3t+2x,4t+3x,t} > > > >> y = 2x + 3t; z = 3x + 4t is a solution to x^2 + 2y^2 = z^2 + 2t^2? > > > >> x^2 + 2(4x^2 + 12xt + 9t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 > > >> x^2 + 8x^2 + 24xt + 18t^2) = 9x^2 + 24xt + 16t^2 + 2t^2 > > > >> Ok, that is correct. So is y = t, z = x a solution. > > > >> You are looking for elements r,s in the ring R = Z[sqr 2] > > >> for which norm(r) = n(r) = n(s). > > >> n(r) = rr* where (n + m.sqr 2)* = n - m.sqr 2 for n,m in Z. > > > >> There should be an infinite sequence of such numbers. > > >> Find some u /= +-1 in R with n(u) = 1. > > >> For example 3 + 2.sqr 2 or 3 - 2.sqr 2. > > >> Then for all j in N, n(r) = n(u^j r). > > > > For the given example > > > x^2 + 2y^2 = z^2 + 2t^2 {x,t} = {2,3} > > > > which other solutions do you propose? > > > For each n in N, calculate > > > r_n = (2 + 3.sqr 2)(3 +- 2.sqr)^n > > > r_n = a_n + b_n.sqr 2 for some a_n, b_n in Z. > > > Set y = a_n, z = b_n.- Masquer le texte des messages précédents - > > > - Afficher le texte des messages précédents - > > Thanks for your answering me. > My particular solution was based on the > following identity: > (a+2b+2c)^2+(2a+b+2c)^2+c^2 > = > a^2 +b^2 +(2a+2b+3c)^2 > > for a=b > c^2 +2(3a+2c)^2 = (4a+3c)^2+2a^2 > > Alain- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Bonjour, I'll present a sort of extension of form x^2 + 2y^2 = z^2 + 2t^2 Playing with squares of same rectangular sums: Ex:(x+2y+2z)^2+(2x+y+2z)^2 and (2x+2y+3z)^2 = 8xy+12xz+12yz modulo(squares) We may write things like (3x+2y+2z)^2+(2x+y+2z)^2+(x+2y+2z)^2+2z^2 = (2x+2y+3z)^2+(2x+2y+z)^2+(2x+y+2z)^2+2x^2 Alain |