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 $x$, $y$ or $z$ that is part of a solution to the Markov Diophantine Equation: $x^{2} + y^{2} + z^{2} = 3xyz$, so through previous equation we get the sequence $1, 2, 5, 13, 29...$.

In particular one can normalize the triples so that $x \leqslant y \leqslant z$. Then if $(x, y, z)$ is a Markov triple then by Vieta jumping so is $(x, y, 3xy-z)$. Beautiful, isn’t?

But how We got the $(x, y, 3xy-z)$? 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 $n$

(with the coefficients being real or complex numbers and $a_{n} \neq 0$) is known by the Fundamental theorem of algebra to have $n$ complex roots $x_{1}, x_{2}, ..., x_{n}$. Vieta’s formulas relate the polynomial’s coefficients {$a_{k}$} to signed sums and products of its roots {$x_{i}$} as follows:

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

By example:

For the polynomial $P(x) = ax^{3} + bx^{2} + cx + d$, roots $x_{1}, x_{2}, x_{3}$ of the equation $P(x) = 0$ satisfy

## Solving problems

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

$1.-$ Let $a$ and $b$ be positive integers so that $ab + 1$ divides $a^{2} + b^{2}$. Prove that $\frac{a^{2}+b^{2}}{ab+1}$ is a perfect square.

Let $k = \frac{a^{2}+b^{2}}{ab+1}$, we consider all the pairs $(a, b)$ non-negative which satisfy equation:

$\frac{a^{2}+b^{2}}{ab+1} = k,$ then $s =$ {$(a, b): a, b \ \ \epsilon \ \ \mathbb{Z}^{+} \wedge \frac{a^{2}+b^2}{ab+1} = k$}

There exists a pair $(a, b)$ where $b=0$, in this case we have:

So, for this case our proof is complete. For prove this claim, we suppose that $k$ is not a perfect square and that $(A, B \ \epsilon \ \ S)$ is the pair that minimize $a+b$ over all pairs $(a, b) \ \epsilon \ \ S$. So, we assume that: $A \geqslant B > 0$, we have the equation:

which is equivalent to:

as a quadratic equation in $X$. We know that $X_{1} = A$ is one root of this equation, by Vieta’s formula the another root of the equation is:

We realize that $X_{2}$ is an integer and $X_{2}\neq0$ since otherwise, $k=B^{2}$ would be a perfect square, contradicting our assumption. Also, $X_{2}$ cannot be negative, for otherwise:

a contradiction, hence, $X_{2} \geqslant 0$ and thus $(a, b) \ \epsilon \ \ S$. However, because $A \geqslant B$, we have:

So $% $, contradicting the minimality of $A+B \ \$ Q.E.D.

$2.-$ (IMO 2007) Let $a, b$ be positive integers. Show that if $4ab-1$ divides $(4a^{2}-1)^{2}$, then $a=b$

Let $k = \frac{(4a^{2}-1)^{2}}{(4ab-1)}$, we consider the pair $(a, b)$ non-negative which satisfy this equation:

$\frac{(4a^{2}-1)^{2}}{(4ab-1)} = k$, Then $s =$ {$(a,b): a, b \ \epsilon \ \ \mathbb{Z}^{+} \wedge \frac{(4a^{2}-1)^{2}}{(4ab-1)} = k$},

There exists a pair $(a, b)$ where $a=b$, in this case we have:

First case:

Let’s make $2a=x$, then $(x^{2}-1) = k$

Second case:

Let’s make $2b=y$, then $y^{2}-1 = k$

Then,

Substituting

So, for this case our proof is complete. For prove this claim, we suppose that $a \neq b$ and that $(A, B) \ \epsilon \ \ s$ is the pair where $a$ and $b$ are never equals to all pairs $(a, b) \ \ \epsilon \ s$. So, we assume that $% B > 0) \vee (0 < A < B) %]]>$

Let’s make $2A = X$

Which is equivalent to say:

As a quadratic equation in $X$. By Vieta’s formula we know that $X_{1}=2A$ is one of roots of this equation, the another one is:

We realize that $X_{2}$ is an integer and $% B) \vee (B$, otherwise $A$ could be equal to $B$, contradicting our assumption. Also we note that for any $k \ \epsilon \ \ \mathbb{Z}^{+}$, Neither $A>B$ or $% $ is not strictly necessary. So,

A contradiction, hence $a \neq b$. So $X_{2} + B \leqslant A + B$, contradicting the statement of $A \neq B$. Q.E.D