1. ## method of Ascent

Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

I don't fully get the concept of the method of Ascent. Could you please show me how ? Thank you very much.

2. Originally Posted by beta12
Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

I don't fully get the concept of the method of Ascent. Could you please show me how ? Thank you very much.
This is the classical diophantine equation from number theory. It is inacurrately called, "Pellian Equation".

Some History.
---
The first first person to make use of it was Fermat in 17th Century. He claimed he had a proof of the infinitude of solutions using the method of infinite descent (not ascent). Not supprsingly, he never published it. Nor do I think there exists a proof using infinite descent.

A century later Euler decided to solve this problem in general. But even he failed. Eventually, Lagrange succedded in showing that:
x^2-ny^2=1
Where n is a non-square integer has infinitely many solutions.

I have only seen two proofs. One which I believe was worked out by Euler/Lagrange by the use of infinite continued fractions. [Every solution is a convergent of the continued fractional expansion of sqrt(n)]. It takes like 6 pages (if you already know the fundamental properties of continued fractions).

The other proof I have only gazed upon. It was done by Dirichelt. Without the use of continued fractions. All he did was show the infininitude of solution (where Euler/Lagrange has an algorithm).

Both the proofs are really ugly/hard/long using non-stop use of inequalities.

But the basic theorem you need to know that, is "n" is NOT a square then there are infinitely many solutions. Otherwise, the only solution is trivial.

3. Hi Perfecthacker,

Thanks for the information.

4. Let me at least solve it for you.
---
We expand sqrt(3) as a continued fraction.

[1; 1,2,1,2,1,2,....]
The length of this expansion is n=2 which is even.
Thus, fundamental solution is:
x=p_{n-1} y=q_{n-1}
Thus,
x=p_1 and y=q_1

We need to compute the convergent C_1 which is,
1+1/1=2/1
Thus,
x=2 and y=1
Which is true.

Thus all solutions are,
[2+sqrt(3)]^n
When you exand it you have an integer and an integer with a radical. The free integer is the x value and the integer with the radical is the y value.

For example, if n=2 then,
[2+sqrt(3)]^2=4+4sqrt(3)+3=7+4sqrt(3)
Thus,
x=7 and y=3

If, n=3 then,
[2+sqrt(3)][7+4sqrt(3)]=26+15sqrt(3)
Thus,
x=26 and y=15

And so on...
~~~
Alternatively, all convergents C_{2k-1} for k=1,2,3,4,...
Are solution to this equation.
This shows why there are infinitely many.

5. Hi Perfecthacker,
Thanks for the solution with continued fractions approach.

I must solve this question with the method of Ascent. I've got some more information for this question. Can you figure out the proof by the method of Ascent?
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Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

You do this by showing how, given one solution (u,v), we can compute another solution(w,z) that is "larger " is some suitable sense. Then your proof will involve finding a pair of formulas , something like:
w = x +y and z = x-y

(These formulas won't work. But there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3)

6. Originally Posted by beta12
Use the method of Ascent to prove there are infinitely many solutions to the Diophantine equation: x^2 - 3y^2 =1.

You do this by showing how, given one solution (u,v), we can compute another solution(w,z) that is "larger " is some suitable sense. Then your proof will involve finding a pair of formulas , something like:
w = x +y and z = x-y

(These formulas won't work. But there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3)
Ah!
I see what you are saying. I was referring to Fermant's method of descent when you show no solutions exist and I believed you misspelled the word. Okay now I understand.

You basically want to have a constructive proof, meaning from one solution to obtain another.
But the trick is to show that at least one solution exists, which is the problem.

I need time to think about it and respond back.

7. Originally Posted by ThePerfectHacker
Ah!
I see what you are saying. I was referring to Fermant's method of descent when you show no solutions exist and I believed you misspelled the word. Okay now I understand.

You basically want to have a constructive proof, meaning from one solution to obtain another.
But the trick is to show that at least one solution exists, which is the problem.

I need time to think about it and respond back.
A simple solution exists: x=2, y=1.

RonL

8. Originally Posted by CaptainBlack
A simple solution exists: x=2, y=1.

RonL
Right

I was thinking when I wrote the post,
x^2-ny^2=1
In general

9. If,
x=a
y=b
Is a solution,
Then,
(a^2+3b^2)^2-3(2ab)^2=(a^2-3b^2)^2=1

Thus an new solution is,
x=a^2+3b^2=A
y=2ab=B

Thus a new solution is,
x=A^2+3B^2
y=2AB

Continue doing this and you shall obtain a newer solution, each being differenct from the others because they form a strictly monotonic increasing sequence.

Thus, we "ascent"

Note: This not necessarily produce all the solutions but it shows there are infinitely many. I think when Lagrange proved this diophantine equation he did not use continued fractions rather he shows the existence of a non-trivial solution and then used this argument to how one can construct more solutions.

10. Hi Perfecthacker,

Thank you very much. I got it now.