# [SOLVED] Solving Diophantine Equations:

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• Jun 7th 2010, 06:23 AM
Samson
[SOLVED] Solving Diophantine Equations:
Hello All, I am trying to parameterize solutions in integers to the following Diophantine Equation:

(x^2) + 5*(y^2) = (z^2)

This comes from an additional review portion of my text. It states the following tip:

Consider only solutions that satisfy GCD[x,y,z]=1, and that x,y,z > 0 and that x is odd. If two cases appear, combine them into one by using absolute value.

Does anybody have any ideas? Any help is appreciated!
• Jun 7th 2010, 06:46 AM
undefined
Quote:

Originally Posted by Samson
Hello All, I am trying to parameterize solutions in integers to the following Diophantine Equation:

(x^2) + 5*(y^2) = (z^2)

This comes from an additional review portion of my text. It states the following tip:

Consider only solutions that satisfy GCD[x,y,z]=1, and that x,y,z > 0 and that x is odd. If two cases appear, combine them into one by using absolute value.

Does anybody have any ideas? Any help is appreciated!

I think you will end up with something similar to generating Pythagorean triples. I'll have to think about it/review some sources. Or maybe someone will post a nice solution in the meantime.
• Jun 7th 2010, 06:53 AM
Samson
Quote:

Originally Posted by undefined
I think you will end up with something similar to generating Pythagorean triples. I'll have to think about it/review some sources. Or maybe someone will post a nice solution in the meantime.

For what it's worth, I know you're correct considering it was in a section of my text that explicitly mentiones Pythagorean triples. I know that doesn't help provide proof, but at least is a good sign that we are in the right ballpark.
• Jun 7th 2010, 07:12 AM
chiph588@
Quote:

Originally Posted by Samson
Hello All, I am trying to parameterize solutions in integers to the following Diophantine Equation:

(x^2) + 5*(y^2) = (z^2)

This comes from an additional review portion of my text. It states the following tip:

Consider only solutions that satisfy GCD[x,y,z]=1, and that x,y,z > 0 and that x is odd. If two cases appear, combine them into one by using absolute value.

Does anybody have any ideas? Any help is appreciated!

I'm stumped for the moment being, but I'd suggest trying to manipulate the equation into the form $a^2+(5y)^2=b^2$ where $a,b$ are in terms of $x,y,z$. There's probably an easier way though.
• Jun 7th 2010, 07:15 AM
roninpro
Taking the idea from the Wikipedia article, I used the following:

$a=m^2-5n^2$, $b=2mn$, $c=m^2+5n^2$

You can put this into the equation to see that it does indeed work. Then I suppose that you can throw in an extra parameter $k$ to get all solutions (with some possible overlap).
• Jun 7th 2010, 07:20 AM
Samson
Quote:

Originally Posted by roninpro
Taking the idea from the Wikipedia article, I used the following:

$a=m^2-5n^2$, $b=2mn$, $c=m^2+5n^2$

You can put this into the equation to see that it does indeed work. Then I suppose that you can throw in an extra parameter $k$ to get all solutions (with some possible overlap).

Okay, I'm just a little confused. I gave my equations with x,y,z and you have a,b,c,m,n, and possibly k... could you relate this back into the terms I had given? What are a,b,c,m,n, and k compared to x,y,z ? Thank you!
• Jun 7th 2010, 07:24 AM
undefined
Quote:

Originally Posted by Samson
Okay, I'm just a little confused. I gave my equations with x,y,z and you have a,b,c,m,n, and possibly k... could you relate this back into the terms I had given? What are a,b,c,m,n, and k compared to x,y,z ? Thank you!

Pythagorean triples are typically given as a^2 + b^2 = c^2, as on the Wikipedia page.

So your (x,y,z) became (a,b,c). m and n are positive integers that we iterate over. k is a multiplier. It is explained in more detail in the Wikipedia article. However, since gcd(x,y,z) = 1, you will not need the multiplier.
• Jun 7th 2010, 07:27 AM
roninpro
Oh, sorry, I meant to put $x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$.

You allow $m, n$ to range freely throughout the integers. Doing this gives you infinitely many solutions. Now, we have a little bit of a problem: we might not catch all of the solutions. To fix this, we throw in a parameter $k$, which also ranges through the integers. This gives

$x=k(m^2-5n^2)$, $y=k(2mn)$, $z=k(m^2+5n^2)$

This should give you every possible solution.
• Jun 7th 2010, 07:27 AM
chiph588@
Quote:

Originally Posted by roninpro
Taking the idea from the Wikipedia article, I used the following:

$a=m^2-5n^2$, $b=2mn$, $c=m^2+5n^2$

You can put this into the equation to see that it does indeed work. Then I suppose that you can throw in an extra parameter $k$ to get all solutions (with some possible overlap).

To prove these are the only solutions, perhaps you could perform a manipulation like I had suggested earlier? From that we know all solutions to a^2+b^2=c^2.
• Jun 7th 2010, 07:33 AM
Samson
Quote:

Originally Posted by roninpro
Oh, sorry, I meant to put $x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$.

You allow $m, n$ to range freely throughout the integers. Doing this gives you infinitely many solutions. Now, we have a little bit of a problem: we might not catch all of the solutions. To fix this, we throw in a parameter $k$, which also ranges through the integers. This gives

$x=k(m^2-5n^2)$, $y=k(2mn)$, $z=k(m^2+5n^2)$

This should give you every possible solution.

Okay, I'm confused by this. Chip says that we don't need k because the GCD[x,y,z]=1. However we need to find all of the solutions that satisfy the conditions I stated originally. How do we do this or have we already? If so, which ones are they?
• Jun 7th 2010, 07:38 AM
roninpro
Quote:

Originally Posted by chiph588@
To prove these are the only solutions, perhaps you could perform a manipulation like I had suggested earlier? From that we know all solutions to a^2+b^2=c^2.

I'm not entirely sure that will work; one solution that I found was $x=2, y=1, z=3$, which gives $4+5=9$ - we have some trouble with the 5.

I think that some insight for the proof comes from Dr. Math: Math Forum: Ask Dr. Math FAQ: Pythagorean Triples
• Jun 7th 2010, 07:45 AM
Samson
Quote:

Originally Posted by roninpro
I'm not entirely sure that will work; one solution that I found was $x=2, y=1, z=3$, which gives $4+5=9$ - we have some trouble with the 5.

I think that some insight for the proof comes from Dr. Math: Math Forum: Ask Dr. Math FAQ: Pythagorean Triples

So if $x=2, y=1, z=3$ and they are the only solutions, why would the formulas below be used?

$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$
• Jun 7th 2010, 07:50 AM
undefined
Quote:

Originally Posted by Samson
So if $x=2, y=1, z=3$ and they are the only solutions, why would the formulas below be used?

$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$

(x,y,z) = (2,1,3) is one of infinitely many solutions.
• Jun 7th 2010, 07:53 AM
roninpro
Quote:

Originally Posted by Samson
So if $x=2, y=1, z=3$ and they are the only solutions, why would the formulas below be used?

$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$

By "only solutions", Chip meant that the solutions to the Diophantine equation have the form given in our equation. And actually, we may have a problem. I don't think that we can represent $2=m^2-5n^2$ for some $m,n$. So, I suppose that some modification to the formula is required.

Maybe you can have a look at that Dr. Math article and try a similar argument.
• Jun 7th 2010, 07:54 AM
Samson
Quote:

Originally Posted by undefined
(x,y,z) = (2,1,3) is one of infinitely many solutions.

So would it be correct to say that

(x,y,z)=(2,1,3) is one solution, but

$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$

is a parameterization of all solutions that meet the aforementioned conditions?
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