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This question is motivated by a remark of Bill Dubuque on my answer to the following question:

An Algorithm to compute the GCD of polynomials of coprime numbers?

Vieta jumping is based on the idea that if you have a quadratic equation with integer coefficients that is symmetric in $x$ and $y$, you can try to find a descent argument by replacing $x$ by the other root of the equation and then changing the roles of $x$ and $y$.

see also

To quote Bill Dubuque's comment:

"Vieta jumping" is a strange name for what is essentially descent in a group of integer points on a conic (realized by reflection). This is a special case of results on Pell equations (or equivalent theories).

And here is finally my question:

Does the Pell equation point of view give a good characterisation when this procedure will actually give a descent?

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It would be very useful if someone could give an example of this "descent". Or describe what it is. – quanta May 11 '11 at 19:57
In the gray question, descent just means that the integers do get smaller, so that the procedure stops eventually. But this should be linked to a degenerate case of descent for Pell equations (or not, that's why I ask this quesiton). So, there are two kinds of descent of my question and the goal of the question is to unify them. – Phira May 11 '11 at 20:07
I really have no idea why this question is related to Viète. As to the desired examples, it is already given in the link. I saw nothing about the equation of Pell which guarantees the validity of the method, not able to be found in the elementary ones. From the point of view of me, it depends largely upon the quadratic equation you constructed to solve the question. Maybe this is what you meant by the characterization? In any case, what is the difficulty in the process of solving real questions, may I know? – awllower Aug 18 '11 at 15:03
up vote 5 down vote accepted

If $(x/z , y/z)$ is a rational point of the Pell conic $X^2 - \Delta Y^2 = 4$ and $gcd(x,z)=1$, then the Diophantine equation $z T^2 + x T U + z U^2 = y^2$ in variables $T, U$ can be viewed as a descendant of the Pell conic, and there is a descent via the jumping you refer to. Two points are required to perform the descent. Fortunately if we have $(T, U)$ then we also have $(U, T)$. One can `subtract' one point from the other. The specific details of how to get this to work are in my Ph.D. thesis, which I intend to submit next month but it follows from the article Arithmetic of Pell surfaces, Acta Arith., (2011).

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The rest of my answer is at: – Samuel Hambleton Oct 9 '11 at 7:29

Old question; the interesting application of Vieta Jumping is in several variables, for example Diophantine quartic equation in four variables, part deux and links there. The fundamental setup is $$ x_1^2 + x_2^2 + \cdots x_n^2 = F(x_1, x_2, \ldots , x_n), $$ where $F$ is a symmetric polynomial for which every term is "squarefree" in each variable,all exponents are just $1.$ So, Markov Numbers are $$ x^2 + y^2 + z^2 = 3 x y z. $$ Integral Apollonian circle packings are $$ w^2 + x^2 + y^2 + z^2 = 2 w x + 2 w y + 2 w z + 2 x y + 2 x z + 2 y z. $$ The right hand side need not be homogeneous. A constant term is permitted, but that tends to throw things off because $(0,0,\ldots,0)$ is no longer a solution. The earlier MSE question with this is $$ w^2 + x^2 + y^2 + z^2 = wxyz -2 w x - 2 w y - 2 w z - 2 x y - 2 x z - 2 y z. $$ Notice how all variables (right hand side) appear without explicit visible exponents, so all exponents are actually $1.$

Anyway, what happens is that you get a forest of solutions, split up into rooted trees, possibly infinitely many. There are infinitely many in the Apollonian thing. Also, there, the roots include some negative numbers. By and large, it appears as though, when the right hand side $F$ has maximum degree higher than $2,$ we can probably split off the positive solutions into a finite number of trees.

EEDDIIIITT, May 24, 2014: This turned out to be interesting, Diophantine quartic equation in four variables asking for all solutions in positive integers to $$ (w + x + y + z)^2 = wxyz. $$ While there are solutions with entries $0$ or negative, it turned out that we may split off the positive solutions, into exactly nine rooted trees. Go Figure.

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Evidently i had this edit as the first sentence. Live and learn. – Will Jagy May 24 '14 at 22:30

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