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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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#1
Dec312, 09:39 AM

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I am trying to find a way to generate solutions to 5b^2*c^2 = 4a^2(b^2+c^2)
Can anyone offer some insight? I know that (b^2+c^2) is the part that is divisible by 5 


#2
Dec312, 10:55 PM

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The interesting cases will be when HCF(a,b,c)=1, so assume that. First, consider what cases that leaves where some two of the three have a common factor.
When no two have a common factor, look for an interesting factorisation. With squares, that's typically going to be of the form (xy)(x+y). Hint: try writing the 5 as 4+1. 


#3
Dec312, 11:09 PM

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#4
Dec412, 12:39 AM

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How to solve for solutions to the diophantine 5b^2*c^2 = 4a^2(b^2+c^2)
I'm suggesting breaking the problem into 3 cases:
1. a, b, c have a common factor. This is trivially reducible to the other cases by factoring it out. 2. Some pair of a, b, c have a common factor. E.g. consider p divides a m times and b n times. Then you can show either p = 2 or m=n, and maybe deduce some more consequences. 3. No two have a common factor. Rewrite the equation as 4b^{2}(c^{2}a^{2}) = c^{2}(4a^{2}b^{2}) then factorise. What can the prime factors of c divide on the LHS? 


#5
Dec412, 12:47 AM

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I know you are trying to help but I sincerely have absolutely no idea what that aims to solve?
It seems like you're advocating some form of just iterating through all a,b,c in order to get all the cases and break them out into these three categories, but I'm not sure how this is any better than brute force. 1. You're saying gcd(a,b,c)>1 here? 2. either gcd(a,b)>1, gcd(a,c)>1, or gcd(b,c)>1? 3. gcd(a,b,c)=1 How does rewriting the equation that way help, and factorize which part? If it helps any, I am only looking for cases for which gcd(a,b,c)=1 and a<b<c<2*a 


#6
Dec412, 12:55 AM

P: 250

for example
[209, 247, 286], [341, 374, 527], [779, 950, 1025], [1711, 2146, 2183] ... etc 


#7
Dec412, 01:13 AM

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Lastly, I just tried looking at the prime factors of c with respect to the lhs and nothing unusual cropped up (for instance using the cases I just posted). It's not like the primes all exclusively divide 4b^2 or (c^2a^2) if that's what you're saying.
It can divide 4*b*b*a*a though; not sure if this matters 


#8
Dec412, 03:10 AM

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No, not brute force. I'm suggesting that in each case you can make interesting but different logical deductions which might eventually allow you to characterise all solutions.
Take e.g. gcd(a,c)=1 and gcd(b,c) = 1. Then c = 1 or 2. Indeed, if gcd(a,b)=1 then bc. 


#9
Dec412, 04:35 AM

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#10
Dec412, 07:22 AM

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I don't know why Dickfore; it just is
haruspex: Yes but I am after the gcd(a,b,c) = 1 cases which don't seem to have those same unique attributes 


#11
Dec412, 08:27 AM

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#12
Dec412, 02:34 PM

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#13
Dec412, 05:40 PM

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Doesn't (b^2 + c^2) or a^2 have to be divisible by 5 as the left side of the equation has 5 as a factor. Also b or c must be even as 5(B^2)(C^2) is equal to 4 times a number.



#14
Dec412, 08:54 PM

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#15
Dec412, 08:57 PM

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#16
Dec412, 09:07 PM

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I've since figured out my own problem, no need to discuss this any further



#17
Dec512, 01:48 AM

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I am always fascinated by solutions to Diophantine equations. I am glad that you figured it out, but we haven't learned anything from this discussion. Can you please share your solution with us? Thanks, LittleNewton 


#18
Dec512, 07:19 AM

P: 894

PS, Although you may have been irritated by one or more of us, please be advised that none of the posts in this thread appear to have been meant to belittle you. It just that its sometimes hard to choose from simply spoon feeding detailed information to someone and just giving that person enough information so that he or she will have the ability to effectively solve a problem. I know that it is human nature for each of us to sometimes have a mental block at times and at that time a little more information to wake us up and help us think straight may be more helpful in the long run than just spoon fed detailed information. 


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