How to solve for solutions to the diophantine 5b^2*c^2 = 4a^2(b^2+c^2)

[ramsey2879]

I did a bit more experiments and found that I can get the range check down to
order of forth root: 2*limit^(0.25) to be precise.

Also using gcd = 1, skips many candidates.
Focusing on signs and parity also cuts it down.

I kept a hash of the results to avoid repeats,
but if I a nice method is outlined there shouldn't be a need
to store the results.

I don't know what you mean by limit. I don't see why using gcd limits candidates should matter since if (d,e) > 1 then GCD(d,e,f) > 1 and I thought SeventhSigma was only interested in primitive solutions. As for limiting d,e to positive, it seems trival to test both d and -d when testing d will suffice, same for e.; but, then maybe you have a program to compute d and e that will give negative input. If so then go for it.

 

[LittleNewton]

I don't know what you mean by limit. I don't see why using gcd limits candidates should matter since if (d,e) > 1 then GCD(d,e,f) > 1 and I thought SeventhSigma was only interested in primitive solutions. As for limiting d,e to positive, it seems trival to test both d and -d when testing d will suffice, same for e.; but, then maybe you have a program to compute d and e that will give negative input. If so then go for it.

I am trying to get the primitive solutions under a certain limit.
The origin of this idea was micromass' suggetion:

Finding all the solutions of these is not very hard. This contains some nice information: http://mathcircle.berkeley.edu/BMC4/Handouts/elliptic/node4.html

I did case (2) explicitely, and the solutions of 4d^2+e^2=5f^2 are given by

d= m^2 - 2mn - 4n^2,~~ e= m^2 + 8mn - 4n^2,~~ f= m^2 + 4n^2

where we can take m and n coprime. We still got to take care that d and e are coprime, since they might not be in some cases. If we want f and e to be odd, then it clearly suffices to ask that m is odd.

In the interest of generating the primitives efficiently, and
I am trying to eliminate the computation of gcd's if I can,
because there are too many degenerate (multiple) cases.
I think the solution would be along the lines of generating a solution
set similar to tree of primitive Pythagorean triples.

 

[ramsey2879]

I edited the last post in page three to note that if d(m,n) = m^2-2mn-4n^2 and that if e(m,n) = m^2+8mn -4n^2 is a solution for f = m^2 + 4*n^2 then so are d(m,-n) and e(m,-n). Moreover, I proved that d(m,n)*e(m,-n) + d(m,-n)*e(m,n) equals 2f^2, that is 2*(m^2+4n^2)^2.
Currently, I am looking at a method of using matrix operations with the set {{d,e},{d'e'}} to get from a solution for f = r to a solution for f=r^2. What is interesting is that the same matrix operation can be repeated n times to go from a solution for f = 1 to a solution for f = r^n. This may be a precurser to a general matrix operation for going from f=r to f = r*s, but this later bit may just be wishful thinking.

 

[ramsey2879]

I edited the last post in page three to note that if d(m,n) = m^2-2mn-4n^2 and that if e(m,n) = m^2+8mn -4n^2 is a solution for f = m^2 + 4*n^2 then so are d(m,-n) and e(m,-n). Moreover, I proved that d(m,n)*e(m,-n) + d(m,-n)*e(m,n) equals 2f^2, that is 2*(m^2+4n^2)^2.
Currently, I am looking at a method of using matrix operations with the set {{d,e},{d'e'}} to get from a solution for f = r to a solution for f=r^2. What is interesting is that the same matrix operation can be repeated n times to go from a solution for f = 1 to a solution for f = r^n. This may be a precurser to a general matrix operation for going from f=r to f = r*s, but this later bit may just be wishful thinking.

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.

 

[ramsey2879]

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]

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?