Vieta jumping


In number theory, Vieta jumping, also known as root flipping, is a proof technique. It is most often used for problems in which a relation between two positive integers is given, along with a statement to prove about its solutions. There are multiple methods of Vieta jumping, all of which involve the common theme of infinite descent by finding new solutions to an equation using Vieta's formulae.

History

Vieta jumping is a relatively new technique in solving mathematical olympiad problems, as the first olympiad problem to use it in a solution was proposed in 1988 for the International Mathematics Olympiad and assumed to be the most difficult problem on the contest. Arthur Engel wrote the following about the problem's difficulty:
Among the eleven students receiving the maximum score for solving this problem were future Fields-medallist Ngô Bảo Châu, Stanford math professor Ravi Vakil, Mills College professor Zvezdelina Stankova, and Romanian politician Nicușor Dan. Emanouil Atanassov, Bulgaria, solved the problem in a paragraph and received a special prize.

Standard Vieta jumping

The concept of standard Vieta jumping is a proof by contradiction, and consists of the following three steps:
  1. Assume toward a contradiction that some solution exists that violates the given requirements.
  2. Take the minimal such solution according to some definition of minimality.
  3. Show that this implies the existence of a smaller solution, hence a contradiction.
; Example
Problem #6 at IMO 1988: Let and be positive integers such that divides. Prove that is a perfect square.
  1. Fix some value that is a non-square positive integer. Assume there exist positive integers for which.
  2. Let be positive integers for which and such that is minimized, and without loss of generality assume.
  3. Fixing, replace with the variable to yield. We know that one root of this equation is. By standard properties of quadratic equations, we know that the other root satisfies and.
  4. The first expression for shows that is an integer, while the second expression implies that since is not a perfect square. From it further follows that is a positive integer. Finally, implies that and thus, which contradicts the minimality of.

    Constant descent Vieta jumping

The method of constant descent Vieta jumping is used when we wish to prove a statement regarding a constant having something to do with the relation between and. Unlike standard Vieta jumping, constant descent is not a proof by contradiction, and it consists of the following four steps:
  1. The equality case is proven so that it may be assumed that.
  2. and are fixed and the expression relating, and is rearranged to form a quadratic with coefficients in terms of and, one of whose roots is. The other root, is determined using Vieta's formulas.
  3. It is shown that for all above a certain base case, and that is an integer. Thus we may replace with and repeat this process until we arrive at the base case.
  4. The statement is proven for the base case, and as has remained constant through this process, this is sufficient to prove the statement for all ordered pairs.
; Example
Let and be positive integers such that divides. Prove that.
  1. If must divide and thus and. So, without loss of generality, assume that.
  2. Let and rearrange and substitute to get. One root to this quadratic is, so by Vieta's formulas the other root may be written as follows:.
  3. The first equation shows that is an integer and the second that it is positive. Because as long as.
  4. The base case we arrive at is the case where. For this to satisfy the given condition, must divide, making either 1 or 2. The first case is eliminated because. In the second case,. As has remained constant throughout this process, this is sufficient to show that will always equal 3.

    Geometric interpretation

Vieta jumping can be described in terms of lattice points on hyperbolas in the first quadrant. The same process of finding smaller roots is used instead to find lower lattice points on a hyperbola while remaining in the first quadrant. The procedure is as follows:
  1. From the given condition we obtain the equation of a family of hyperbolas that are unchanged by switching and so that they are symmetric about the line.
  2. Prove the desired statement for the intersections of the hyperbolas and the line.
  3. Assume there is some lattice point on some hyperbola and without loss of generality. Then by Vieta's formulas, there is a corresponding lattice point with the same -coordinate on the other branch of the hyperbola, and by reflection through a new point on the original branch of the hyperbola is obtained.
  4. It is shown that this process produces lower points on the same branch and can be repeated until some condition is achieved. Then by substitution of this condition into the equation of the hyperbola, the desired conclusion will be proven.
; Example
This method can be applied to problem #6 at IMO 1988: Let and be positive integers such that divides. Prove that is a perfect square.
  1. Let and fix the value of. Then represents a lattice point on the hyperbola defined by the equation.
  2. If then we find, which trivially satisfies the statement.
  3. Let be a lattice point on a branch, and assume so that it is on the higher branch. By applying Vieta's Formulas, is a lattice point on the lower branch of. Then, by reflection is a lattice point on the original branch. This new point has smaller -coordinate, and thus is below the original point. Since this point is on the upper branch, it is still above.
  4. This process can be repeated. From the equation of, it is not possible for this process to move into the second quadrant. Thus, this process must terminate at and by substitution, is a square as required.