# 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
 P: 250 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
 Homework Sci Advisor HW Helper Thanks P: 9,920 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 (x-y)(x+y). Hint: try writing the 5 as 4+1.
P: 250
 Quote by haruspex 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 (x-y)(x+y). Hint: try writing the 5 as 4+1.
I am sorry, I don't really understand what you mean here at all. Can you provide an example?

 Homework Sci Advisor HW Helper Thanks P: 9,920 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 4b2(c2-a2) = c2(4a2-b2) then factorise. What can the prime factors of c divide on the LHS?
 P: 250 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
 P: 250 for example [209, 247, 286], [341, 374, 527], [779, 950, 1025], [1711, 2146, 2183] ... etc
 P: 250 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^2-a^2) if that's what you're saying. It can divide 4*b*b*a*a though; not sure if this matters
 Homework Sci Advisor HW Helper Thanks P: 9,920 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 b|c.
P: 3,014
 Quote by SeventhSigma I know that (b^2+c^2) is the part that is divisible by 5
Why?
 P: 250 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
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P: 18,346
 Quote by SeventhSigma I don't know why Dickfore; it just is
That doesn't really make any sense. If you don't know why, then why did you add the condition?? Is it given in the problem statement that 5 divides $a^2+b^2$?? Or...
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P: 9,920
 Quote by SeventhSigma I don't know why Dickfore; it just is
Well, that is very easy to prove.
 haruspex: Yes but I am after the gcd(a,b,c) = 1 cases which don't seem to have those same unique attributes
Yes, I understand that. The cases I elaborated upon, like (a,b)=1, necessarily satisfy (a,b,c)=1. Do you mean you are most interested in the cases where (a,b,c)=1 but some pair does have a common factor? Or perhaps where each pair has a common factor?
 P: 234 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.
P: 3,014
 Quote by haruspex Well, that is very easy to prove.
Apparently, it isn't for the OP. If one is incapable of verifying this statement, then the analysis you suggested is beyond their comprehension.
P: 3,014
 Quote by coolul007 Doesn't (b^2 + c^2) or a^2 have to be divisible by 5?
So, what if we assume that $b^2 + c^2$ is not divisible by 5?
 P: 250 I've since figured out my own problem, no need to discuss this any further
P: 12
 Quote by SeventhSigma I've since figured out my own problem, no need to discuss this any further
Hi SeventhSigma,

I am always fascinated by solutions to Diophantine equations.
I am glad that you figured it out, but we haven't learned anything