Vieta jumping, explained simply: the jump and when to reach for it

Vieta jumping has a reputation it does not deserve. The mechanic is four lines of school algebra, and every one of them is something you learned before you learned what an olympiad was: a quadratic has two roots, their sum is $-b/a$, their product is $c/a$.

What is genuinely hard is the two decisions wrapped around those four lines. Knowing that a problem wants this at all, and knowing what to do when the jump lands somewhere awkward. So that is what this is: the trigger, the jump explained simply enough to reproduce under exam pressure, and the case that quietly costs marks.

The shape that triggers it

Vieta jumping is not a general-purpose tool. It fires on a narrow shape, and if you can recognise the shape you have done most of the work.

• Two positive-integer unknowns, and the condition is symmetric in them. Swapping $a$ and $b$ has to leave the problem unchanged. The symmetry is what lets you always jump the larger one.
• The condition is quadratic in each variable separately. Vieta's formulas need a quadratic. If either variable appears cubed, this is not your technique.
• There is a quantity you can freeze. Almost always a quotient or a value $k$ that the problem either names or asks you to describe. You will fix $k$ and treat it as a constant for the rest of the argument.
• Modular arithmetic does not kill it. If you can rule the equation out mod something, do that instead; it is shorter. Vieta jumping is for the problems where solutions genuinely exist and you have to say something structural about them.

The jump, in four lines

Take the archetype. If $ab+1$ divides $a^2+b^2$, the quotient is a perfect square. Write $k$ for that quotient and clear the denominator:

$$a^2 + b^2 = k(ab+1).$$

Now the four lines.

1. Freeze the quotient and read the equation as a quadratic. Treat $k$ as a constant, move everything to one side and look at it as a polynomial in $a$ alone, with $b$ and $k$ as constants: $a^2 - (kb)a + (b^2-k) = 0$.
2. Name the other root. Call it $a'$. Vieta says $a + a' = kb$ and $a\,a' = b^2 - k$.
3. The sum makes the new root an integer. $a' = kb - a$ is a difference of integers, so no rationals sneak in. This is the whole trick, and it is why the technique is named after Vieta rather than after the descent.
4. The product makes the new root small. $a' = (b^2-k)/a$, so if $a > b$ then $a' < b^2/a < b$. The pair $(a', b)$ solves the same equation with the same $k$ and a strictly smaller sum.

That last line is the jump. You started at a solution and landed on a smaller one, and the two together are a machine that cannot run forever inside the positive integers. Which is infinite descent wearing a quadratic disguise: it is not a different idea, it is the same idea with Vieta supplying the smaller object for free.

The case that loses marks

Step 4 said $a'$ is smaller. It did not say $a'$ is positive, and the descent only contradicts anything if the new pair is a genuine solution in positive integers. Skipping this is the single most common way a correct-looking Vieta write-up drops points.

There are two cases and you must handle both.

A negative second root is impossible. Substitute it back. With $a' \le -1$, $b \ge 1$ and $k \ge 1$, the expression $a'^2 - kb\,a' + b^2 - k$ is at least $1 + kb + b^2 - k = 1 + k(b-1) + b^2$, which is strictly positive. A root cannot make the polynomial positive, so this case never happens.

A second root of zero is where the descent stops, and it is the answer. The product formula gives $a\,a' = b^2 - k$, so $a' = 0$ forces $k = b^2$. The chain does not run forever because it terminates on exactly the conclusion you were asked to prove. The perfect square was not something you engineered; it fell out of the only exit the descent has.

That is worth sitting with, because it is the pattern in every good descent proof. You are not proving "no solutions" here, you are proving that every solution slides down to a base case, and then reading what the base case says.

Where the solutions actually live

Run the machine backwards and you can see the whole solution set, which is the thing that makes this problem beautiful rather than merely hard.

Start from a base pair $(n, n^3)$. It works for every $n$: the quotient is $(n^6+n^2)/(n^4+1) = n^2$. Then climb instead of descending, replacing $b$ by $ka - b$ each time. With $k=4$ the base is $(2,8)$ and the ladder is $2, 8, 30, 112, 418$, each term four times the previous minus the one before it.

So the solutions are not scattered. They are a family of ladders, one per square $k$, each rooted at $(n, n^3)$. And they are sparse: over every pair with $1 \le b \le a \le 400$, only ten satisfy the divisibility at all. That sparsity is why searching for a pattern by hand gets you nowhere on this problem, and why almost nobody solves it by experiment.

How it differs from plain descent

Compare it with a straight descent problem: $x^2 + y^2 = 3z^2$ has no positive integer solutions. There the smaller solution comes from divisibility. Squares are $0$ or $1$ mod $3$, which forces $3$ to divide both $x$ and $y$, and dividing through produces a smaller triple. The same argument in its oldest form proves $\sqrt2$ is irrational, where the shrinking is a common factor of $2$ you can never finish removing.

The difference is only in where the smaller object comes from.

• Plain descent: the smaller solution comes from a divisibility fact. You show a prime divides everything, then divide.
• Vieta jumping: the smaller solution comes from an algebraic fact. The quadratic already has a second root, and Vieta hands it to you.

Both end the same way, on the impossibility of an infinite decreasing sequence of positive integers. If you have understood one you are three-quarters of the way to the other, which is the argument for learning them as a pair rather than as two tricks.

What to do next

Do not go looking for Vieta jumping problems immediately. There are not many, and hunting them teaches you the shape without teaching you the descent underneath it. Do the plain descent problems first until the phrase "take the smallest counterexample" is automatic, then come back and the quadratic will look like a convenience rather than a new technique.

Lemma treats these as one named technique for exactly that reason. The infinite descent and Vieta jumping page has the statement and where it sits among the others, the problem archive carries the problems above in full with difficulty and expected time on each, and the daily problem posts a new one every day in three tiers with no account needed.