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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
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Dickfore
#19
Dec5-12, 09:46 AM
P: 3,014
Quote Quote by SeventhSigma View Post
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
#20
Dec5-12, 09:57 AM
P: 250
Quote Quote by ramsey2879 View Post
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
#21
Dec5-12, 12:13 PM
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Quote Quote by SeventhSigma View Post
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 [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.
SeventhSigma
#22
Dec5-12, 12:21 PM
P: 250
Quote Quote by micromass View Post
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 [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.
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
#23
Dec5-12, 01:16 PM
P: 12
Quote Quote by SeventhSigma View Post
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.

Quote Quote by SeventhSigma View Post
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:

Quote Quote by SeventhSigma
Quote Quote by LittleNewton
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
#24
Dec5-12, 01:43 PM
P: 894
Quote Quote by SeventhSigma View Post
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
#25
Dec5-12, 01:49 PM
P: 250
Quote Quote by ramsey2879 View Post
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
#26
Dec5-12, 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
micromass
#27
Dec5-12, 02:01 PM
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Quote Quote by SeventhSigma View Post
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
#28
Dec5-12, 03:17 PM
P: 12
Quote Quote by SeventhSigma View Post
LittleNewton is pretty much doing what I am doing though
May be you can help me to finish up my solution.
dodo
#29
Dec5-12, 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; so there *has* to be another way to obtain a multiple of 5 other than from this particular pythagorean triple.

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
#30
Dec6-12, 02:04 PM
P: 894
Quote Quote by LittleNewton View Post
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
#31
Dec6-12, 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.
micromass
#32
Dec6-12, 04:36 PM
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Quote Quote by LittleNewton View Post
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/...tic/node4.html

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.
LittleNewton
#33
Dec6-12, 04:55 PM
P: 12
Quote Quote by micromass View Post
Finding all the solutions of these is not very hard. This contains some nice information: http://mathcircle.berkeley.edu/BMC4/...tic/node4.html

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.
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
#34
Dec6-12, 05:19 PM
P: 250
Hint:

Reduce the cases you know are correct to their primitives

Check which bounds produce those primitives
LittleNewton
#35
Dec6-12, 05:39 PM
P: 12
Quote Quote by SeventhSigma View Post
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
#36
Dec6-12, 07:05 PM
P: 12
Quote Quote by LittleNewton View Post
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...


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