# Diophantine equation question

• Oct 19th 2010, 04:44 PM
Brimley
Diophantine equation question
I've read through a few exercises on Number Theory with regards to Diophantine Equations. Apparently, some of them are solvable and some are not. Here is the example that conflicts me:

Are there solutions to the Diophantine equation $x^2 + y^2 = 7*z^2$ ? If so, what are they?
• Oct 19th 2010, 06:20 PM
undefined
Quote:

Originally Posted by Brimley
I've read through a few exercises on Number Theory with regards to Diophantine Equations. Apparently, some of them are solvable and some are not. Here is the example that conflicts me:

Are there solutions to the Diophantine equation $x^2 + y^2 = 7*z^2$ ? If so, what are they?

I don't have time to work on this, but you may find this thread helpful

http://www.mathhelpforum.com/math-he...ns-148115.html

And the proof technique of this post (#22) might apply

http://www.mathhelpforum.com/math-he...tml#post525708
• Oct 19th 2010, 09:02 PM
Brimley
undefined, I read that thread you had mentioned and the Wiki articles as well, however those threads had a lot of differing techniques and it was for a specific sample that involved the GCD of the variables equalling 1. I guess what I"m getting at was it looked like you, roninpro, and chip had many different approaches for the same problem, and I tried applying the proof technique of post #22 but didn't get anywhere.

If you or someone else does have a little time to spare, I would really like to see if there are any sols to this and if so find out what they are.

Much thanks for any help on this! - Brim
• Oct 19th 2010, 09:19 PM
undefined
Quote:

Originally Posted by Brimley
undefined, I read that thread you had mentioned and the Wiki articles as well, however those threads had a lot of differing techniques and it was for a specific sample that involved the GCD of the variables equalling 1. I guess what I"m getting at was it looked like you, roninpro, and chip had many different approaches for the same problem, and I tried applying the proof technique of post #22 but didn't get anywhere.

If you or someone else does have a little time to spare, I would really like to see if there are any sols to this and if so find out what they are.

Much thanks for any help on this! - Brim

I will try to find time but can't make promises. Off hand though I can tell you that trivially yes there is a solution, (x,y,z) = (0,0,0).
• Oct 19th 2010, 09:27 PM
Unbeatable0
It's simpler to use Fermat's infinite descent:

Show that the only solution to $x^2+y^2\equiv\: 0\: (\text{mod} \ 7)$ is $x,y\equiv\: 0\: (\text{mod} \ 7)$

Then suppose $x,y,z$ are solutions to the original problem. From above we get:

$49\left(\frac{x}{7}\right)^2+49\left(\frac{y}{7}\r ight)^2=7z^2$

Therefore $7\mid z$ and we can write:

$49\left(\frac{x}{7}\right)^2+49\left(\frac{y}{7}\r ight)^2=7^3\left(\frac{z}{7}\right)^2$

$\left(\frac{x}{7}\right)^2+\left(\frac{y}{7}\right )^2=7\left(\frac{z}{7}\right)^2$

Therefore if $(x,y,z)$ is a solution then so is $\left(\frac{x}{7},\frac{y}{7},\frac{z}{7}\right)$, and applying again and again:

$\left(\frac{x}{7^n},\frac{y}{7^n},\frac{z}{7^n}\ri ght)$

is a soloution for all $n\in\mathbb{N}$, and in particular made of integers.

But it's impossible that these numbers are integers for all $n$ if one of $x,y,z$ is nonzero, so we have the only solution: $(0,0,0)$
• Oct 19th 2010, 09:45 PM
Brimley
Quote:

Originally Posted by Unbeatable0
Therefore if $(x,y,z)$ is a solution then so is $\left(\frac{x}{7},\frac{y}{7},\frac{z}{7}\right)$, and applying again and again:

$\left(\frac{x}{7^n},\frac{y}{7^n},\frac{z}{7^n}\ri ght)$

is a soloution for all $n\in\mathbb{N}$, and in particular made of integers.

But it's impossible that these numbers are integers for all $n$ if one of $x,y,z$ is nonzero, so we have the only solution: $(0,0,0)$

1. What does the $\mathbb{N}$ mean in $n\in\mathbb{N}$ ?
2. So the only solution is actually the trivial solution of $(0,0,0)$[/QUOTE] ?
• Oct 20th 2010, 04:38 AM
Unbeatable0
Quote:

Originally Posted by Brimley
1. What does the $\mathbb{N}$ mean in $n\in\mathbb{N}$ ?
2. So the only solution is actually the trivial solution of $(0,0,0)$

1. It means the set of natural numbers (integer numbers greater than or equal to 1)
2. Indeed
• Oct 20th 2010, 08:59 AM
Unbeatable0
We can generalize this problem:

Consider the diophantine equation $ax^2+by^2=cz^2$ with $a,b\in\mathbb{Z}\:,\:c\in\mathbb{N}$ satisfying the conditions:

1) $q$, the "squarefree part" of $c$ (that is, $c=m^2q$ where $q$ is squarefree) is odd.

2) $a,b$ are coprime to $q$.

3) $\left(\frac{ab}{q}\right) = (-1)^\frac{q+1}{2}$ (Jacobi Symbol)

then the only solution to the diophantine equation is the trivial solution.

Note that in the special case of $ab=\pm 1$, (2) is obviously true and we can replace condition (3) with:

$(3)_{\pm1}\:\:q \equiv2+ab\pmod{4}$

(the question which started the thread is an obvious special case of the above)

Proof:

$ax^2+by^2=cz^2 \Leftrightarrow ax^2+by^2=q(mz)^2$

so it's enough to prove that $ax^2+by^2=qz^2$ has no nontrivial solutions.

Let $(x,y,z)$ be a solution. Then

$ax^2+by^2 \equiv 0 \pmod{q} \Rightarrow x^2\equiv -a^{-1}by^2 \pmod{q}$.

Suppose that $\gcd(y,q)\ne 1$, and let $r\mid \gcd(y,q)$ be a prime number. Using the same idea of infinite descent as in my first post in this thread, we get a contradiction.

Thus $\left(\frac{y}{q}\right)\ne 0 \Rightarrow \left(\frac{y^2}{q}\right) = 1$ and therefore:

$\left(\frac{-a^{-1}by^2}{q}\right) = \left(\frac{-a^{-1}by^2}{q}\right)\left(\frac{a^2}{q}\right) = \left(\frac{-ab}{q}\right)\left(\frac{y^2}{q}\right) = \left(\frac{-1}{q}\right)\left(\frac{ab}{q}\right)$

$=(-1)^\frac{q-1}{2}(-1)^\frac{q+1}{2} = -1$

which contradicts $x^2\equiv -a^{-1}by^2 \pmod{q}$. Therefore $q\mid x$. By symmetry, $q\mid y$, and we can write:

$aq^2\left(\frac{x}{q}\right)^2+bq^2\left(\frac{y}{ q}\right)^2 = qz^2$.

(here we refer to the usual fractions, not Jacobi symbols)

Dividing by $q$ and using the fact that it's squarefree we get $q\mid z$. Then

$\left(\frac{x}{q},\frac{y}{q},\frac{z}{q}\right)$

is another solution to the diophantine equation. Continuing ad infinitum yields a contradiction if not all $x,y,z$ are zero.

We thus get the desired result.
• Oct 20th 2010, 09:24 AM
Brimley
Thank you very much unbeatable, i thoroughly enjoyed reading your proof. Thanks again -- Brim