How to solve for solutions to the diophantine 5b^2*c^2 = 4a^2(b^2+c^2)by SeventhSigma Tags: 4a2b2, 5b2c2, diophantine, solutions, solve 

#55
Dec1212, 10:18 AM

P: 891

{{1,1},{1,1}}*r' equals {{x1,y1},{x2,y2}} where (x1,y1) and (x2,y2) are solution sets for a^2*x^2 + b^2*y^2 = (a^2+b^2)*N^2 {{1,1},{1,1}}*r'*r' equals {{x3,y3},{x4,y4}} ... are solution sets for a^2*x^2+b^2y^2 = (a^2+b^2)*N^4 {{1,1},{1,1}}*r'*r'*r' = {{x5,y5},{x6,y6}} ... are solution sets for a^2*x^2 + b^2*y^2 = (a^2+b^2)*N^6 etc. 



#56
Dec1312, 06:26 AM

P: 891

[QUOTE=ramsey2879;4194504]I generalized the above solution. Given a,b,m,n, then r' = {{a*a*n*nm*m,2*a*a*m*n/b},{2*m*n*b,a*a*n*nm*m}} . Now
{{1,1},{1,1}}*r' equals {{x1,y1},{x2,y2}} where (x1,y1) and (x2,y2) are solution sets for a^2*x^2 + b^2*y^2 = (a^2+b^2)*N^2 {{1,1},{1,1}}*r'*r' equals {{x3,y3},{x4,y4}} ... are solution sets for a^2*x^2+b^2y^2 = (a^2+b^2)*N^4 {{1,1},{1,1}}*r'*r'*r' = {{x5,y5},{x6,y6}} ... are solution sets for a^2*x^2 + b^2*y^2 = (a^2+b^2)*N^6 etc.[/QUOTEI] I forgot to ask the obvious questions, I know that for a=4,b = 1 the above formulas give all solution sets except for reordering and sign. Is this true for all a and b? If not how can it be modified or what needs to be added? 


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