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]

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?

 

[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]

I know that (b^2+c^2) is the part that is divisible by 5

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]

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...

 

[haruspex]

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?

 

[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]

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.

 

[Dickfore]

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?

 

[SeventhSigma]

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

 

[LittleNewton]

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

Thanks,

LittleNewton

 

[ramsey2879]

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

Well it would be helpful to us to give us some insight as to what you discovered or found. As to the proof that (b^2 + c^2) must be divisible by 5 Dickfore was trying to get you to think like a mathematican. If 5 doesn't divide (b^2 + c^2) then it must divide a^2. But if 5 divides a^2 and not b^2 + c^2 then the right hand side must be divisible by an even power of 5. Is ths so with the left hand side? Don't let this forum get you down, it is very helpful for those willing to think for themselves. It would be nice for you to return the favor.
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.

 

[Dickfore]

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

So, could you post your solution here for others to use?

 

[SeventhSigma]

Well it would be helpful to us to give us some insight as to what you discovered or found. As to the proof that (b^2 + c^2) must be divisible by 5 Dickfore was trying to get you to think like a mathematican. If 5 doesn't divide (b^2 + c^2) then it must divide a^2. But if 5 divides a^2 and not b^2 + c^2 then the right hand side must be divisible by an even power of 5. Is ths so with the left hand side? Don't let this forum get you down, it is very helpful for those willing to think for themselves. It would be nice for you to return the favor.
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.

I made the reply "I don't know, it just is" when I was really tired and cranky. I did know why and what I meant by that was "I don't know, just look at it dude, it's trivial to prove."

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.

 

[micromass]

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.

This is very unfair! Especially haruspex made some very helpful suggestions to you!

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 b^2+c^2 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.

 

[SeventhSigma]

This is very unfair! Especially haruspex made some very helpful suggestions to you!

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 b^2+c^2 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.

It was something I posted in the very first post in this thread. I didn't want to waste time proving trivial things (it felt like someone was asking me to prove 2+2=4 or something; it's just obvious from looking at it), since I am not interested in proving a bunch of stuff. I just wanted to surge ahead to understanding a solution.

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.

 

[LittleNewton]

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.

After we solve a problem, all hard things look simple and trivial. I can't blame haruspex for that.

Everyone else was just being rude for no reason.

I think this is punishing all, for the behavior of the few...

And it looks like SeventhSigma doesn't want to share his solution in any case:

In case you don't decide to share your solution publicly,
please at least reply to me.

not going to happen, sorry

I believe the purpose of any forum is to have a 2 way communication,
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.

(d-f)*(d+f) = 4*(f-e)*(f+e) -> ((d-f)/2)*((d+f)/2) = (f-e)*(f+e)

but I don't know how to go on to finding all the initial seeds and
moving on to infinitely many solutions.

 

[ramsey2879]

Everyone else was just being rude for no reason.

Not true, nothing in your first or later posts gave us any hint as to what your level of understanding was. In fact you mislead us, so how were we to know what help you didn't need. This problem remains interesting with or without your continued input. Sorry you have such a inclination to belittle our input.

 

[SeventhSigma]

Not true, nothing in your first or later posts gave us any hint as to what your level of understanding was. In fact you mislead us, so how were we to know what help you didn't need. This problem remains interesting with or without your continued input. Sorry you have such a inclination to belittle our input.

Absolutely nothing I said was remotely misleading; I said from the very start that b^2+c^2 was divisible by 5.

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.

 

[SeventhSigma]

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

 

[micromass]

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

Are you going to tell us your solution or not??

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.

 

[LittleNewton]

LittleNewton is pretty much doing what I am doing though

May be you can help me to finish up my solution.

 

[dodo]

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; [strike]so there *has* to be another way to obtain a multiple of 5 other than from this particular pythagorean triple.[/strike]

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.

 

[ramsey2879]

After we solve a problem, all hard things look simple and trivial. I can't blame haruspex for that.
I think this is punishing all, for the behavior of the few...

And it looks like SeventhSigma doesn't want to share his solution in any case:
I believe the purpose of any forum is to have a 2 way communication,
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.

(d-f)*(d+f) = 4*(f-e)*(f+e) -> ((d-f)/2)*((d+f)/2) = (f-e)*(f+e)

but I don't know how to go on to finding all the initial seeds and
moving on to infinitely many solutions.

I did a program and came to the following conjecture; If and only If "f" is an product of primes of the form 4n+1, does 4*d^2 + e^2 = 5*f^2 have a solution such that d and e are coprime. Could anyone verify this. Note that primes (5, 13 etc.) are each considered a product of primes. In other words, in searching for a solution where f is odd, one only search for f being a product of primes of the form 4n+1.
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.

 

[dodo]

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.

 

[micromass]

After we solve a problem, all hard things look simple and trivial. I can't blame haruspex for that.



I think this is punishing all, for the behavior of the few...

And it looks like SeventhSigma doesn't want to share his solution in any case:



I believe the purpose of any forum is to have a 2 way communication,
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.

(d-f)*(d+f) = 4*(f-e)*(f+e) -> ((d-f)/2)*((d+f)/2) = (f-e)*(f+e)

but I don't know how to go on to finding all the initial seeds and
moving on to infinitely many solutions.

Finding all the solutions of these is not very hard. This contains some nice information: http://mathcircle.berkeley.edu/BMC4/Handouts/elliptic/node4.html

I did case (2) explicitely, and the solutions of 4d^2+e^2=5f^2 are given by

d= m^2 - 2mn - 4n^2,~~ e= m^2 + 8mn - 4n^2,~~ f= m^2 + 4n^2

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.

 

[LittleNewton]

Finding all the solutions of these is not very hard. This contains some nice information: http://mathcircle.berkeley.edu/BMC4/Handouts/elliptic/node4.html

I did case (2) explicitely, and the solutions of 4d^2+e^2=5f^2 are given by

d= m^2 - 2mn - 4n^2,~~ e= m^2 + 8mn - 4n^2,~~ f= m^2 + 4n^2

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.

Yes, I tried something similar to that, but it produces a lot of overlaps.

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.

 

[SeventhSigma]

Hint:

Reduce the cases you know are correct to their primitives

Check which bounds produce those primitives

 

[LittleNewton]

Hint:

Reduce the cases you know are correct to their primitives

Check which bounds produce those primitives

How can I judge? Can you be more specific?

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

 

[LittleNewton]

How can I judge? Can you be more specific?

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

By trial and error, I found that the cube root of limit is enough for the range.
However this might be by chance or might depend on that particular limit.

And that is still a lot of (primitive) starting points...

 

[LittleNewton]

Finding all the solutions of these is not very hard. This contains some nice information: http://mathcircle.berkeley.edu/BMC4/Handouts/elliptic/node4.html

I did case (2) explicitely, and the solutions of 4d^2+e^2=5f^2 are given by

d= m^2 - 2mn - 4n^2,~~ e= m^2 + 8mn - 4n^2,~~ f= m^2 + 4n^2

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.

I also read a lot of papers on that subject.
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 non-existence of a solution.

I would like to know when we are sure that we have found all the primitives.

 

[micromass]

I also read a lot of papers on that subject.
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 non-existence of a solution.

I would like to know when we are sure that we have found all the primitives.

I'm pretty sure that those are all the solutions. Did you check the link in my post? There they give a proof (or at least an argument) for why these are all the solutions.

 

[SeventhSigma]

By trial and error, I found that the cube root of limit is enough for the range.
However this might be by chance or might depend on that particular limit.

And that is still a lot of (primitive) starting points...

false

 

[micromass]

Hint:

Reduce the cases you know are correct to their primitives

Check which bounds produce those primitives

Please don't post unless you post your solution.

 

[LittleNewton]

I'm pretty sure that those are all the solutions. Did you check the link in my post? There they give a proof (or at least an argument) for why these are all the solutions.

Yes, I checked it. Somehow only they (this paper and others)
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?

 

[micromass]

Wow, that was fast!
My reply also got deleted...

I'm sorry about this thread. I should have locked it when I had the chance. Now I let it degenerate into a mess. My sincere apologies to all the readers.

 

[micromass]

Yes, I checked it. Somehow only they (this paper and others)
only confidently talk about primitives in case of the basic Pythagorean triples.

I was always under the impression that we could find all the rational points on a given conic. I'll admit that my knowledge about this is very small. But in the book "Rational points on elliptic curves" by Silverman and Tate, they certainly seem to indicate that all the rational points on a conic can be found with this method.

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

 

[LittleNewton]

I was always under the impression that we could find all the rational points on a given conic. I'll admit that my knowledge about this is very small. But in the book "Rational points on elliptic curves" by Silverman and Tate, they certainly seem to indicate that all the rational points on a conic can be found with this method.

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

I guess I am not very good at this either
(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 non-primitive 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 case-specific.

 

[LittleNewton]

I guess I am not very good at this either
(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 non-primitive 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 case-specific.

It might be that the only problem is efficiency:
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...

 

[ramsey2879]

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.

My conjecture is stronger than that since 9*13 is an even power of 3 times a prime of the form 4n +1. 5*9 and 5*113^2 cannot be put into the form 4n^2+m^2 with n coprime to m [even] though 9 and 81 are even powers of 3

 

[ramsey2879]

I also read a lot of papers on that subject.
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 non-existence of a solution.

I would like to know when we are sure that we have found all the primitives.

My experience with Mathematica at this problem leads me to conjecture that the number of primitives is 1 if f is a power of 5, 2 if a power of a prime of the form 4n + 1 or 5 times a power of such a prime, for a product of n distinct primes of the form 4n + 1 or 5 times such a product, the number of primitives is 2^n, If f containes a factor of the form 4n+3 there is no primitive root.
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.

 

[ramsey2879]

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^2-2mn-4n^2) + p*(m^2+8mn -4n^2) = 2*(m^2+4n^2)^2.

Substitute q =b'(m,-n) =m^2-8mn-4n^2 and p = a'(m,-n) =m^2+2mn-4n*2
then (m^2-8mn-4n^2)*(m^2-2mn-4n^2) + (m^2+2mn-4n^2)*(m^2+8mn-4n^2) =

m^4 -10m^3*n+(16-8)m^2*n^2+40n^3*m+16n^4 + m^4 +10m^3*n +
(16-8)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.

 

[ramsey2879]

false

This was in response to LittleNewton saying f^3 appears to be an upper limit for searching for primitive solutions.

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.

 

[LittleNewton]

By trial and error, I found that the cube root of limit is enough for the range.
However this might be by chance or might depend on that particular limit.

And that is still a lot of (primitive) starting points...

I did a bit more experiments and found that I can get the range check down to
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.