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How to solve for solutions to the diophantine 5b^2*c^2 = 4a^2(b^2+c^2) 
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#37
Dec612, 07:09 PM

P: 12

Most just state that these generating functions generate infinitely many solutions, but they are not all the solutions, meaning there might be more. They sometimes use modular arithmetic to show the nonexistence of a solution. I would like to know when we are sure that we have found all the primitives. 


#38
Dec612, 07:19 PM

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#39
Dec612, 07:28 PM

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#41
Dec612, 07:43 PM

P: 12

only confidently talk about primitives in case of the basic Pythagorean triples. I want to know : 1) if I am supposed to extent to negative m,n values 2) what ranges of m & n will give me all the primitives For example if I want to find all solutions below a certain limit, say 10K, what is the complete set of primitives I need to start with? 


#42
Dec612, 07:58 PM

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#43
Dec612, 08:01 PM

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Can you explain me what makes this situation different from the Pythagorean triple situation? 


#44
Dec612, 08:18 PM

P: 12

(otherwise I would have found the solution by now ) I think they are both the same. I want to know where to stop finding the primitives and jump to finding the all other nonprimitive solutions. If the answer is that m,n should span 0..limit, then that is a really big range. I want to know if this is the most efficient way, i.e. we can't avoid the big search. I tried to trim it by jumping on evens for m, and on odds for n, etc. but each case missed some solutions. I tried squareroot(limit), which worked. I tried cuberoot(limit), seemed to work, but I guess it was casespecific. 


#45
Dec612, 08:27 PM

P: 12

How can we efficiently generate all solutions? i.e. how can we avoid spending time on degenerate solutions, because it seems like there are lots of them... 


#46
Dec712, 12:43 AM

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#47
Dec712, 09:40 AM

P: 894

My Mathematical program (for odd f )is the following: t = 1; tx = t + 401; While[t < tx, Print[t, Solve[ 4*x*x + y*y == 5*t*t && GCD[x, y] == 1 && x > 0 && y > 0, {x, y}, Integers]]; t += 4] for investigating a specific product like t = 13*17*37*29 I set tx = 1. 


#48
Dec812, 10:30 AM

P: 894

I have a further observation: If 4x^2+y^2=f has at least one primitive solution and 5 does not divide f, then there are two primitive solutions for f (not divisible by 5) in 4x^2 + y^2 = 5f, (x1,y1) and (x2,y2) all positive values such that x2*y1 +/ x1*y2 = +/ 2f^2). Never mind my post about f being prime, f = 29*17 is a counter example to that conjecture.
Below is a proof; The pell equation 4a^2+b^2 = 5f^2 has the solution set a = m^2 2mn 4n^2, b = m^2 +8mn 4n^2, f = m^2 + 4n^2. If (m,n) form one solution set then (m,n) will form a 2nd solution set (p,q). Also, my experience is that qa + pb = 2*f^2. We need to show that q*(m^22mn4n^2) + p*(m^2+8mn 4n^2) = 2*(m^2+4n^2)^2. Substitute q =b'(m,n) =m^28mn4n^2 and p = a'(m,n) =m^2+2mn4n*2 then (m^28mn4n^2)*(m^22mn4n^2) + (m^2+2mn4n^2)*(m^2+8mn4n^2) = m^4 10m^3*n+(168)m^2*n^2+40n^3*m+16n^4 + m^4 +10m^3*n + (168)m^2*n^2  40n^3*m + 16n^4 = 2*(m^4 + 8m^2*n^2 + 16n^4) = 2*(m^2+4n^2). Q.E.D. 


#49
Dec812, 11:13 AM

P: 894

If you have some information on the upper range for searching for primitive solutions being the cube of f then would you provide it? I am not sure what you have in mind but am interested. 


#50
Dec812, 12:47 PM

P: 12

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. 


#51
Dec812, 02:46 PM

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#52
Dec812, 03:19 PM

P: 12

The origin of this idea was micromass' suggetion: 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. 


#53
Dec812, 10:42 PM

P: 894

I edited the last post in page three to note that if d(m,n) = m^22mn4n^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. 


#54
Dec1012, 10:11 AM

P: 894

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^2m^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. 


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