1. ## diophantine problem

Suppose $n$ is a fixed integer (positive or negative) $\neq 0$ and suppose

$x_0^2 - 3y_0^2 = n$

with $x_0 \geq 0,\ y_0 \geq 0$. Let $x_1 = 2x_0 +3y_0,\ y_1=x_0+2y_0$. Show that

$x_1^2-3y_1^2=n,\ \ \ \ \ x_1>x_0\geq 0,\ \ \ y_1>y_0\geq 0$.

Conclude that the quation $x^2-3y^2=n$ either has no solutions or infinitely many solutions.

All I can think of to do is plug in the given values of $x_1$ and $y_1$ to show that they come out to the same equation. No idea what to do after that or even if that is correct to do at all.

2. Hello,

Substitute $x_1$ and $y_1$ in $x_1^2-3y_1^2$ and prove it is equal to n if $x_0^2-3y_0^2=n$.

This proves that if $\{x_0,y_0\}$ is solution, then $\{x_1,y_1\}$ is solution too.

If there is no solution, we're done.
If there is one solution, $\{x_0,y_0\}$, and therefore $\{x_1,y_1\}$ is solution too. Since $x_1^2-3y_1^2=n$, we can prove the same way as above that $\{x_2=2x_1+3y_0 ~,~ y_2=y_1+2y_1 \}$ is solution too. And so on...

Isn't it the infinite ascent ?

So either there is no solution, either there's an infinity of solutions.

3. so then it would be
$x_1^2 - 3y_1^2 = n$ with $x_1 = 2x_0 +3y_0,\ y_1=x_0+2y_0$
$(2x_0 + 3y_0)^2 - 3(x_0 + 2y_0)^2 = n$
$(4x_0^2 + 12x_0y_0 + 9y_0^2) - 3(x_0^2 + 4x_0y_0 + 4y_0^2) = n$
$x_0^2 - 3 y_0^2 = n$
which is equivalent to $x_1^2 - 3y_1^2 = n$

and then from there you say that $x_{n+1}=2x_n+3y_n,\ y_{n+1}=x_n+2y_n$ is also a solution and so forth, so infinite solutions?

4. Originally Posted by Proggy
so then it would be
$x_1^2 - 3y_1^2 = n$ with $x_1 = 2x_0 +3y_0,\ y_1=x_0+2y_0$
$(2x_0 + 3y_0)^2 - 3(x_0 + 2y_0)^2 = n$
$(4x_0^2 + 12x_0y_0 + 9y_0^2) - 3(x_0^2 + 4x_0y_0 + 4y_0^2) = n$
$x_0^2 - 3 y_0^2 = n$
which is equivalent to $x_1^2 - 3y_1^2 = n$

and then from there you say that $x_{n+1}=2x_n+3y_n,\ y_{n+1}=x_n+2y_n$ is also a solution and so forth, so infinite solutions?
Yes, because every time you substitute these expressions, you get back to the initial equation !

5. would changing the 3 in the equation to be a positive integer d that is not a perfect square change the outcome in a significant way?

6. Originally Posted by Proggy
would changing the 3 in the equation to be a positive integer d that is not a perfect square change the outcome in a significant way?
Yes, because it's not sure you'd find an expression in terms of $x_n$ and $y_n$ for $x_{n+1}$ and $y_{n+1}$ such that it yields the initial equation !

7. hmmm, just to make sure I was clear in what I was asking, the original problem would be restated as:

Suppose $n$ is a fixed integer $\neq 0$ and d is a positive integer not a perfect square, and suppose

$x_0^2 - dy_0^2 = n$

with $x_0 \geq 0,\ y_0 \geq 0$. Let $x_1 = ax_0 + bdy_0,\ y_1 = bx_0 + ay_0$, where a>0, b>0 provides a solution to the Fermat-Pell equation, $a^2-db^2=1$. Show that

$x_1^2-dy_1^2=n,\ \ \ \ \ x_1>x_0\geq 0,\ \ \ y_1>y_0\geq 0$.

Conclude that the quation $x^2-dy^2=n$ either has no solutions or infinitely many solutions.

8. what I get:

$x_1^2 - dy_1^2 = n$ with $x_1 = ax_0 + bdy_0,\ y_1=bx_0+ay_0$
$(ax_0 + bdy_0)^2 - d(bx_0 + ay_0)^2 = n$
$(a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2) - d(b^2x_0^2 + abx_0y_0 + a^2y_0^2) = n$
$a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2 - b^2dx_0^2 - abdx_0y_0 - a^2dy_0^2 = n$
$a^2(x_0^2 - dy_0^2) - b^2d(x_0^2 - dy_0^2) = n$
$(a^2 - db^2)(x_0^2 - dy_0^2) = n$

i am just not sure what to do from here

9. Originally Posted by Proggy
hmmm, just to make sure I was clear in what I was asking, the original problem would be restated as:

Suppose $n$ is a fixed integer $\neq 0$ and d is a positive integer not a perfect square, and suppose

$x_0^2 - dy_0^2 = n$

with $x_0 \geq 0,\ y_0 \geq 0$. Let $x_1 = ax_0 + bdy_0,\ y_1 = bx_0 + ay_0$, where a>0, b>0 provides a solution to the Fermat-Pell equation, $a^2-db^2=1$. Show that

$x_1^2-dy_1^2=n,\ \ \ \ \ x_1>x_0\geq 0,\ \ \ y_1>y_0\geq 0$.

Conclude that the quation $x^2-dy^2=n$ either has no solutions or infinitely many solutions.
Oh ok

Then it's exactly the same reasoning. I didn't know about the Fermat-Pell stuff

10. Originally Posted by Proggy
what I get:

$x_1^2 - dy_1^2 = n$ with $x_1 = ax_0 + bdy_0,\ y_1=bx_0+ay_0$
$(ax_0 + bdy_0)^2 - d(bx_0 + ay_0)^2 = n$
$(a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2) - d(b^2x_0^2 + abx_0y_0 + a^2y_0^2) = n$
$a^2x_0^2 + abdx_0y_0 + b^2d^2y_0^2 - b^2dx_0^2 - abdx_0y_0 - a^2dy_0^2 = n$
$a^2(x_0^2 - dy_0^2) - b^2d(x_0^2 - dy_0^2) = n$
$(a^2 - db^2)(x_0^2 - dy_0^2) = n$

i am just not sure what to do from here
heh, your comment on fermat-pell made me notice that they are saying that $(a^2 - db^2)=1$ so
$(a^2 - db^2)(x_0^2 - dy_0^2) = n$
becomes
$1(x_0^2 - dy_0^2) = n$
which is the same and therefore has infinite solutions! i think?

11. Originally Posted by Proggy
heh, your comment on fermat-pell made me notice that they are saying that $(a^2 - db^2)=1$ so
$(a^2 - db^2)(x_0^2 - dy_0^2) = n$
becomes
$1(x_0^2 - dy_0^2) = n$
which is the same and therefore has infinite solutions! i think?
Yes !
I thought you used it (I didn't see your resolution)

But it's not good starting with $x_1^2 - dy_1^2 = n$
Start with $x_1^2 - dy_1^2$ and then prove it equals n. (in fact, remove all the =n at the end of each line ^^)

12. got it, thanks, and sorry for the trick question by not including the pell-fermat part. I barely noticed that extra wording in the problem and it turned out to be the deal breaker! Thanks so much for your help. One more problem to go and my homework for my number thoery class will be over.

13. Another question along these lines is what would the next three ascent be in the solution if the original equation was changed to:
x^2 - 3y^2 = 6