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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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#19
Dec512, 09:46 AM

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#20
Dec512, 09:57 AM

P: 250

And I feel little incentive to share my solution since I feel like I did not get any help from this thread at all whatsoever, but instead unwarranted belittling. 


#21
Dec512, 12:13 PM

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P: 18,017

I can see how some replied here might appear belittling to you, but that really wasn't the intention. We don't know your current level of mathematics, so we don't know what kind of post we should make in order to help you in the best way. So we asked you to motivate why 5 divides [itex]b^2+c^2[/itex] to see if you could do that and to get you thinking like a mathematician. If you say now that we did not help at all, then I consider this quite insulting, especially to haruspex. 


#22
Dec512, 12:21 PM

P: 250

And to be fair, while haruspex is great, he did not say anything I technically did not already know (nor was what he mentioned directly relevant to the solution); it's just that I did not understand what he was saying at the time because I assumed he actually had a solution and was giving hints. Everyone else was just being rude for no reason. 


#23
Dec512, 01:16 PM

P: 12

And it looks like SeventhSigma doesn't want to share his solution in any case: not just take from it, but sometimes give back. So although not much, here is what I have found: I simplified the whole equation so that (a,b,c)=1. Then named the mutual gcd's as d,e,f and the original equation boiled down to 2 cases: 1) d^2 + e^2 = 5*f^2 (d,e odd, and f even) 2) 4*d^2 + e^2 = 5*f^2 (e,f odd) At this point I can use some tricks to simplify each case further, e.g. (df)*(d+f) = 4*(fe)*(f+e) > ((df)/2)*((d+f)/2) = (fe)*(f+e) but I don't know how to go on to finding all the initial seeds and moving on to infinitely many solutions. 


#24
Dec512, 01:43 PM

P: 894




#25
Dec512, 01:49 PM

P: 250

And I'm not "belittling your input." I'm expressing disdain for the fact that I had a question, only to get no closer to the solution but met with a bunch of rude statements because I didn't want to spend time proving something trivial. 


#26
Dec512, 01:53 PM

P: 250

Besides, my "solution" isn't even all that ideal. It's still very, very slow because I don't yet have good ways to limit the variables. So I do have a solution but it's still not THAT much better than brute force. It's just faster than brute force.
LittleNewton is pretty much doing what I am doing though 


#27
Dec512, 02:01 PM

Mentor
P: 18,017

I'm leaving this thread open for other people who might be interested in dicussing this problem. If you got nothing meaningful to add, then please don't post. 


#28
Dec512, 03:17 PM

P: 12




#29
Dec512, 04:31 PM

P: 688

I was wondering if the fact that b^2 + c^2 is divisible by 5, implies that b=3k and c=4k for some k. In other words, I am asking if a square can be the sum of two squares in more than one way.
Edit: Actually, substituting b=3k and c=4k leads, I think, to a contradiction; Edit 2: Nah, b^2 + c^2 cannot be a square, for a similar reasoning as when proving that it had to be a multiple of 5. Forget what I just said. 


#30
Dec612, 02:04 PM

P: 894

Edit: In fact, it appears that for a coprime solution, both d and e must each be prime; but I checked too little of the results to make that a into a conjecture. 


#31
Dec612, 04:25 PM

P: 688

Hi, ramsey2879,
what you found is a consequence of a theorem that says that a number can be expressed as a sum of squares if and only if its prime factors congruent to 3 mod 4 appear with an even power in the factorization. For example, 2x5x7 is not expressible as a sum of squares, 2x5x7x7 is (490 = 7^2 + 21^2), 2x5x7x7x7 is not, and so on. 


#32
Dec612, 04:36 PM

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P: 18,017

I did case (2) explicitely, and the solutions of [itex]4d^2+e^2=5f^2[/itex] are given by [tex]d= m^2  2mn  4n^2,~~ e= m^2 + 8mn  4n^2,~~ f= m^2 + 4n^2[/tex] 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. 


#33
Dec612, 04:55 PM

P: 12

Let's say we want to find all values below 10K. Since we got to this point by simplifying the very original equation, getting back to that set requires to multiply by every integer, after we find the primitives. And I am not sure what range of m,n I should try to get those primitives. Because of the 5, I thought about modulus 5, and tried to run over all 25 cases : 0..5 x 0..5. But then extended it to 121 cases: 5..5 x 5..5 Finally I tried 0..10^4 x 0..10^4. And every time I got new results. It looks like the more I extend it, the more results I am getting. 


#34
Dec612, 05:19 PM

P: 250

Hint:
Reduce the cases you know are correct to their primitives Check which bounds produce those primitives 


#35
Dec612, 05:39 PM

P: 12

I simplified all results using the gcd's. then I tried randomly picking primitives, and all seemed legitimate... 


#36
Dec612, 07:05 PM

P: 12

However this might be by chance or might depend on that particular limit. And that is still a lot of (primitive) starting points... 


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