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.
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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.
Let me at least solve it for you.
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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...
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Alternatively, all convergents C_{2k-1} for k=1,2,3,4,...
Are solution to this equation.
This shows why there are infinitely many.
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)
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.
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.