Overview

Vieta Jumping (also called root flipping) is an advanced technique where we perform fermats's infinite descent with the usual goal of showing there exists finite solutions to a diophantine equations however with a twist where the descent is performed on the roots of a quadriatic expression where the roots are usually the solutions to the equation. The roots are found by vieta and hence the name 'Vieta Jumping'.

Key Idea

First we have a diophantine equation which is symmetric and we want to perform the vita jump on it. Let us represent this by $f(a,b)=k$ with degree 2. Degree 1 is left out as that will result a linear diophantine equation which can be solved by standard methods. Now we can form a quadriatic equation with either $a$ or $b$ as the subject and $k$ as the value of the diophantine equation.

One such example is shown below

We know that any quadriatic equation has 2 roots. Hence we know there is another solution to the equation we constructed before. Let the equations subject be represented by $t$ where the roots of it are $a$ and $x$.

Clearly $x$ is also a value that satisfies $f(x,b)=k$. A important step is to make sure that $x < a$, $x \in \mathbb{Z}$ and that $x>0$ which can be proven using simple inequalities(We find the value of $x$ via vieta's relations). Else we cannot perform a descent.

Okay we got $(x,b)$ as a solution, now what? well we can again make the quadriatic equation but this time with $b$ as the subject! and there will be no problems since the equation is symmetric in the first place. This is sufficient to show a descent and after that we continue with the standard fermat's descent to find the solutions, prove the solution set is finite or for contradiction. (The most popular is contradiction, which is outlined below)

Steps to Reproduce

Here is the standard structure of a Vieta Jumping argument (This is only for reference and not for forced memorisation):

1. Define $k$. Let $k = f(a,b)$ where $f$ is the expression asserted to be an integer (or a specific value).
2. Fix $k$, take the minimal pair. Among all pairs $(a, b)$ of positive integers with $f(a, b) = k$, choose the one minimizing $a + b$ (or just $a$ after WLOG $a \geq b$).
3. Form the quadratic. Rearrange the equation into a quadratic in $a$: the existing value $a$ is one root.
4. Produce the second root. Let $x$ be the other root. By Vieta's sum formula, $x = (\text{something}) - a \in \mathbb{Z}$.
5. Show $x > 0$. Use the product formula $xa = (\text{constant})$. If $k$ is not a perfect square or if the constant is positive, show $x \neq 0$; rule out $x < 0$ separately.
6. Show $x < a$. Use the assumption $a \geq b$ and the product formula: $x = \frac{b^2 - k}{a} \leq \frac{a^2 - k}{a} < a$.
7. Conclude. The pair $(x, b)$ satisfies the same relation with $x + b < a + b$, contradicting minimality. Hence no such pair exists (or $k$ must equal the forced value).

Worked Examples

Example 1 (IMO 1988 Problem 6). Let $a, b$ be positive integers such that $ab + 1 \mid a^2 + b^2$. Prove that $\dfrac{a^2 + b^2}{ab + 1}$ is a perfect square.

Proof. Let $k = \dfrac{a^2+b^2}{ab+1}$. We want to show $k$ is always a perfect square.

Suppose for contradiction that $k$ is a fixed positive integer that is not a perfect square, and that there exists at least one pair $(a,b)$ of non-negative integers with $\frac{a^2+b^2}{ab+1} = k$. Pick the pair $(a, b)$ with $a \geq b \geq 0$ that minimizes $a$.

Rewrite the equation as $a^2 + b^2 = k(ab + 1)$, then rearrange into a quadratic in $a$:

The value $a$ is one root. Let $x$ be the other root. By Vieta's formulas:

• Sum: $x + a = kb$, so $x = kb - a \in \mathbb{Z}$.
• Product: $x \cdot a = b^2 - k$, so $x = \dfrac{b^2 - k}{a}$.

Since $x$ is an integer satisfying the same quadratic, the pair $(x, b)$ satisfies $x^2 - kbx + (b^2 - k) = 0$, which means $x^2 + b^2 = k(xb + 1)$, i.e., $\frac{x^2 + b^2}{xb + 1} = k$ — so $(x, b)$ is also a valid solution, provided $x \geq 0$ and $xb + 1 > 0$.

$x \neq 0$: If $x = 0$, then $xa = b^2 - k = 0$, so $k = b^2$, contradicting our assumption that $k$ is not a perfect square.

$x \geq 0$: Suppose $x < 0$. Then $x^2 + b^2 = k(xb + 1)$. The left side is positive. We need $xb + 1 > 0$, i.e., $x > -1/b$ — but $x$ is a negative integer, so $x \leq -1$, giving $xb \leq -b \leq 0$, so $xb + 1 \leq 1$. Combined with $x^2 + b^2 > 0$, we need $xb + 1 > 0$, i.e., $xb \geq 0$. Since $x < 0$ and $b \geq 1$, this is impossible. So $x \geq 0$, and since $x \neq 0$, we have $x > 0$.

$x < a$: Since $a \geq b \geq 1$:

We need to show this is less than $a$. Note $k \geq 1$, so $b^2 - k \leq b^2 - 1 < b^2 \leq a^2$, giving $x = \frac{b^2-k}{a} < \frac{a^2}{a} = a$.

But then $(x, b)$ is a valid pair achieving the same $k$, with $x < a$ — contradicting the minimality of $a$. Therefore, no non-perfect-square value of $k$ is achievable, and $k$ must be a perfect square. $\blacksquare$

Example 2. Let $a, b$ be positive integers with $ab \mid a^2 + b^2 + 1$. Show that $\dfrac{a^2 + b^2 + 1}{ab} = 3$.

Proof. Let $k = \frac{a^2+b^2+1}{ab}$. First handle the diagonal: if $a = b$, then $a^2 \mid 2a^2 + 1$, so $a^2 \mid 1$, forcing $a = b = 1$ and $k = 3$. Now assume $a > b \geq 1$ and pick the minimal-sum pair.

Rearrange as a quadratic in $a$:

The value $a$ is one root; let $x$ be the other. By Vieta's:

• Sum: $x = kb - a \in \mathbb{Z}$.
• Product: $xa = b^2 + 1 > 0$.

Since $xa = b^2 + 1 > 0$ and $a > 0$, we have $x > 0$.

$x < a$: Since $a > b$:

So $(x, b)$ is a valid pair with $x < a$, contradicting minimality — unless no such pair with $a > b$ exists. The descent terminates when $a = b$, which forces $k = 3$ as shown above. Therefore $k = 3$ for all valid pairs. $\blacksquare$

Common Pitfalls

Forgetting to prove $x > 0$. The extremal argument only produces a contradiction if the new solution is a valid pair of positive integers. Always prove $x > 0$, or handle $x = 0$ as an explicit base case.

Skipping $a = b$. The bound $x < a$ uses $a > b$. The equality case $a = b$ must always be checked first, this often forces the answer immediately.

Not verifying the new pair satisfies the relation. Check that $(x, b)$ (not just $(a, b)$) satisfies the original equation before claiming it's a valid smaller solution.

Negative variables. The extremal principle requires $a + b$ to be bounded below. If the problem allows negative integers, the descent has no floor. Establish that any solution can be transformed to a positive one before applying the method.

Practice Problems