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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
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ramsey2879
#55
Dec12-12, 10:18 AM
P: 894
Quote Quote by ramsey2879 View Post
Since the two solution sets for f= r form a matrix {{a,b},{p,q}} and the solution set for f=1 is {{1,-1},{1,1}} I wondered what

if we did repeated matrix operation with a matrix r' = r/1 so that 1*r' = r, r*r' = the solution set for f=r^2 etc. Thus r' = {{4n^2-m^2, 8mn},{-2mn, 4n^2 - m^2}}.

As an example, subsitute m=1,n=2: r' = {{15,16},{-4,15}}.

{{1,-1},{1,1}}*r' ={{19,1},{11,31}}

{{19,1},{11,31}}*r'={{281,319},{41,641}}

{{281,319},{41,641}*r' ={{2939,9281},{-1949,10271}}

{{2939,9281},{-1949,10271}}*r' ={{6961,186239},{-70319,122881}}

etc.

These are the solution sets for f =17, 17^2, 17^3, 17^4, etc. Note that the matrix {{1,-1},{1,1} can be changed to any sets long as there are 3 of one sign, and 1 of the other, the same matrix operations will give the solution sets for powers of 17, only the format, i.e. the order or signs,etc. will be changed. However, you can't mess with the solution sets for positive powers of a number or the matrix operation will likely give out garbage.
I generalized the above solution. Given a,b,m,n, then r' = {{a*a*n*n-m*m,2*a*a*m*n/b},{-2*m*n*b,a*a*n*n-m*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.
ramsey2879
#56
Dec13-12, 06:26 AM
P: 894
[QUOTE=ramsey2879;4194504]I generalized the above solution. Given a,b,m,n, then r' = {{a*a*n*n-m*m,2*a*a*m*n/b},{-2*m*n*b,a*a*n*n-m*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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