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

SeventhSigma

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

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.

SeventhSigma

I am sorry, I don't really understand what you mean here at all. Can you provide an example?

haruspex

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?

SeventhSigma

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

SeventhSigma

for example

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

SeventhSigma

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

haruspex

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.

Dickfore

Why?

SeventhSigma

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

micromass

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 [itex]a^2+b^2[/itex]?? Or...

haruspex

Well, that is very easy to prove.
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?

coolul007

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.

Dickfore

Apparently, it isn't for the OP. If one is incapable of verifying this statement, then the analysis you suggested is beyond their comprehension.

Dickfore

So, what if we assume that [itex]b^2 + c^2[/itex] is not divisible by 5?

SeventhSigma

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

LittleNewton

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
from this discussion. Can you please share your solution with us?

Thanks,

LittleNewton