# Thread: yet another Diophantine equation

1. ## yet another Diophantine equation

Hi again,

Have struggled with this for a while but getting nowhere.

$x^2=y^2-2yz-z^2$

I've reduced this to

$x^2=(y-z)^2-2z^2$

but can't seem to break it down any further in a similar method to my previous thread

http://www.mathhelpforum.com/math-he...tml#post674932

Any ideas would be greatly appreciated

2. ## Re: yet another Diophantine equation

Originally Posted by procyon
Hi again,

Have struggled with this for a while but getting nowhere.

$x^2=y^2-2yz-z^2$

I've reduced this to

$x^2=(y-z)^2-2z^2$

but can't seem to break it down any further in a similar method to my previous thread

http://www.mathhelpforum.com/math-he...tml#post674932

Any ideas would be greatly appreciated
Put $w=y-z$ then you have:

$w^2-x^2=2z^2$

CB

3. ## Re: yet another Diophantine equation

Originally Posted by CaptainBlack
Put $w=y-z$ then you have:

$w^2-x^2=2z^2$

CB
Thanks for the reply Captain. I probably should have added that for clarity but thought since I showed

$this^2=that^2-2(theOther)^2$

would be enough.

It's a way of finding integer roots of this (not all, but not by trial & error either) I'm trying to get, similar to the last post I mentioned

http://www.mathhelpforum.com/math-he...tml#post674932

Pro

5. ## Re: yet another Diophantine equation

Originally Posted by procyon
There are no integer solutions when $z$ is odd, to see this (and how to find the other solutions) write:

$w^2-x^2=(w+x)(w-x)=2z^2$

Then for any factorisation $z_1z_2$ of $z^2$ we may set

$w+x=2z_1$

$w-x=z_2$

If $z_2$ is odd then from the second of these equations $w$ and $x$ are of different parities, but from the first they are of the same parity. This is a contradiction so for there to a solution $z_2$ must be even, which is not possible if $z$ (and also $z^2$) is odd.

CB

6. ## Re: yet another Diophantine equation

Many thanks for the above CB. I had been working at it a more difficult way as usual, lol.

I'm not sure if I have made myself clear on what it is I actually seek.

I want to be able to reduce

$w^2-x^2=2z^2$

into some form of

$a^2+b^2=c^2$

even if it is like this

$\left(\dfrac{d+e}{f}\right)^2+\left(\dfrac{g+h}{i} \right)^2=\left(\dfrac{j+k}{l}\right)^2$

I'll start here $x^2+2z^2=w^2$

I have used the fact that z must be even and that w and x must both be even or both be odd to no avail

i.e.

$z$ is even and so there is some integer $p>0$ such that $z=2p$

let w and x both be even, so similarly, $w=2q$ and $x=2r$

so we now have $4r^2+8p^2=4q^2 \Leftrightarrow q^2-r^2=2p^2$

which is back to the start

let's forget that z must be even but still have w and z such

$4r^2+2z^2=4q^2\Leftrightarrow 2r^2+z^2=2q^2$

It's different, but still not what I'm after.

7. ## Re: yet another Diophantine equation

Originally Posted by CaptainBlack
There are no integer solutions when $z$ is odd, to see this (and how to find the other solutions) write:

$w^2-x^2=(w+x)(w-x)=2z^2$

Then for any factorisation $z_1z_2$ of $z$ we may set

$w+x=2z_1$

$w-x=z_2$

If $z_2$ is odd then from the second of these equations $w$ and $x$ are of different parities, but from the first they are of the same parity. This is a contradiction so for there to a solution $z_2$ must be even, which is not possible if $z$ is odd.

CB
Surely, you mean:

$w+x=2z_1^2$ ---------- (1)

$w-x=z_2^2$ ---------- (2)

------------------------------------------------------------------------------------------

Let $z_1=1$ and $z_2=2$

$z=z_1z_2=2$

Solving (1) and (2), we get

$w=3$ and $x=-1$

$y-z=3$

$y-2=3$

$y=5$

So, (-1, 5, 2) is a solution.

Note that (1, 5, 2) is also a solution. (This is obtained by setting $w+x=z_2^2$ and $w-x=2z_1^2$.)

We can choose any $z_1$, any even $z_2$ and generate a solution. Thus, there are infinitely many solutions.

8. ## Re: yet another Diophantine equation

Where did the $y$ appear from?

9. ## Re: yet another Diophantine equation

Originally Posted by procyon
Where did the $y$ appear from?
$w=y-z$, remember?

10. ## Re: yet another Diophantine equation

Originally Posted by alexmahone
$w=y-z$, remember?
I've got that bogged down in this that I'd forgotten, lol. Sorry.

11. ## Re: yet another Diophantine equation

Thank you so much to alexmahone and CaptainBlack for helping me with this.

Here's the whole thing in case it's of use to someone else.
(I've used p and q for the factors only because I find it easier to follow)

$(y-z)^2-x^2=2z^2$

so $\{(y-z)+x\}\{(y-z)-x\}=2z^2$

let $p$ and $q$ be factors of $z$ so that $z=pq$ and $2z^2=2p^2q^2$

now $\{(y-z)+x\}\{(y-z)-x\}=2p^2q^2$

let $\ \ (y-z)+x=2p^2$..........(1)

and $\ \ (y-z)-x=q^2$..........(2)

$(1)+(2)\ \ \ \ \ y-z=\dfrac{2p^2+q^2}{2}\ \ \ \ and\ \ \ \ y=\dfrac{2p^2+2pq+q^2}{2}$

$(1)-(2)\ \ \ \ \ x=\dfrac{2p^2-q^2}{2}$

and now for the checking...

$(y-z)^2-x^2=\left(\dfrac{2p^2+q^2}{2}\right)^2-\left(\dfrac{2p^2-q^2}{2}\right)^2=\dfrac{8p^2q^2}{4}=2p^2q^2$

$and\ \ \ 2z^2=2(pq)^2=2p^2q^2$

Thanks again guys

Pro

12. ## Re: yet another Diophantine equation

Originally Posted by alexmahone
Surely, you mean:

$w+x=2z_1^2$ ---------- (1)

$w-x=z_2^2$ ---------- (2)
You are right that there is a typo, but what I actually meant was $z_1z_2=z^2$, that is that $z_1$ is a factor of $z^2$ and $z_2=z^2/z_1$.

CB

13. ## Re: yet another Diophantine equation

Originally Posted by CaptainBlack
You are right that there is a typo, but what I actually meant was $z_1z_2=z^2$, that is that $z_1$ is a factor of $z^2$ and $z_2=z^2/z_1$.

CB
In that case, I think your method is better than mine.

There is no reason why $w-x$ must be a perfect square or $w+x$ must be twice a perfect square as long as $(w+x)(w-x)$ turns out to be twice a perfect square.

So, my method clearly leaves out some solutions.