Hello guys, this post is about a nice number theory proof technique useful in math contest like IMO, IMC, etc. called the Vieta Root Jumping. The Vieta jumping, which involve the common theme of a proof by infinite descent, by example thanks to Vieta jumping, we can get a Markov tree, which is generated by Markov numbers

Markov number

A Markov number is a positive integer , or that is part of a solution to the Markov Diophantine Equation: , so through previous equation we get the sequence .

In particular one can normalize the triples so that . Then if is a Markov triple then by Vieta jumping so is . Beautiful, isn’t?

But how We got the ? Well, we can do this through Vieta’s formulas

Vieta’s formula

Vieta’s formulas are formulas that relate the coefficients of a polynomial to sums and products of its roots. Any general polynomial of degree

(with the coefficients being real or complex numbers and ) is known by the Fundamental theorem of algebra to have complex roots . Vieta’s formulas relate the polynomial’s coefficients {} to signed sums and products of its roots {} as follows:

Equivalently stated the ()th coefficient is related to a signed sum of all possible subproducts of roots, taken -at-a-time:

By example:

For the polynomial , roots of the equation satisfy

Solving problems

Already we have the basic definitions to prove some statements, so lets begin with the classical problem from IMO 1998:

Let and be positive integers so that divides . Prove that is a perfect square.

Let , we consider all the pairs non-negative which satisfy equation:

then {}

There exists a pair where , in this case we have:

So, for this case our proof is complete. For prove this claim, we suppose that is not a perfect square and that is the pair that minimize over all pairs . So, we assume that: , we have the equation:

which is equivalent to:

as a quadratic equation in . We know that is one root of this equation, by Vieta’s formula the another root of the equation is:

We realize that is an integer and since otherwise, would be a perfect square, contradicting our assumption. Also, cannot be negative, for otherwise:

a contradiction, hence, and thus . However, because , we have:

So , contradicting the minimality of Q.E.D.

(IMO 2007) Let be positive integers. Show that if divides , then

Let , we consider the pair non-negative which satisfy this equation:

, Then {},

There exists a pair where , in this case we have:

First case:

Let’s make , then

Second case:

Let’s make , then

Then,

Substituting

So, for this case our proof is complete. For prove this claim, we suppose that and that is the pair where and are never equals to all pairs . So, we assume that

Let’s make

Which is equivalent to say:

As a quadratic equation in . By Vieta’s formula we know that is one of roots of this equation, the another one is:

We realize that is an integer and , otherwise could be equal to , contradicting our assumption. Also we note that for any , Neither or is not strictly necessary. So,

A contradiction, hence . So , contradicting the statement of . Q.E.D