Vieta Jumping

Warm-up

1. 1
   First verify $(1,1)$: $1 + 1 + 1 = 3 = 3 \cdot 1 \cdot 1$. ✓
2. 2
   Fix $b = 1$ and view the equation as a quadratic in $a$: $a^2 - 3a + 2 = 0$. One root is the known $a = 1$; by Vieta the roots sum to $3$, so the partner root is $a' = 3 - 1 = 2$. Thus $(2, 1)$ is a new solution: $4 + 1 + 1 = 6 = 3 \cdot 2 \cdot 1$. ✓
3. 3
   Now fix $a = 2$ and jump on $b$: $b^2 - 6b + 5 = 0$, with known root $b = 1$ and partner $b' = 6 - 1 = 5$. So $(2, 5)$ works: $4 + 25 + 1 = 30 = 3 \cdot 2 \cdot 5$. ✓
4. 4
   The jumps walk up a ladder $(1,1) \to (2,1) \to (2,5) \to (13,5) \to \cdots$ — exactly the odd-indexed Fibonacci pairs.

Contest level

1. 1
   Let $k = \frac{a^2 + b^2 + 1}{ab}$, a positive integer. Among all positive-integer pairs giving this same $k$, choose one with $a + b$ minimal, and assume $a \ge b$ (WLOG).
2. 2
   Fix $b$ and treat $a^2 - (kb)a + (b^2 + 1) = 0$ as a quadratic in $a$. The other root is $a' = kb - a$ (an integer) with $a a' = b^2 + 1 > 0$, so $a' > 0$ and $(a', b)$ is another valid solution for the same $k$.
3. 3
   By minimality $a' \ge a$. Then $a^2 \le a a' = b^2 + 1 \le a^2 + 1$, and since $a \ge b$ this squeezes $a = b$ (if $a > b$ then $b^2 + 1 \le (a-1)\cdot a < a^2$, contradiction unless $b = a$).
4. 4
   Set $a = b$ in the equation: $2a^2 + 1 = k a^2$, so $a^2(k - 2) = 1$, forcing $a = 1$ and $k - 2 = 1$, i.e. $k = 3$. (Base check: $(1,1)$ gives $\frac{3}{1} = 3$.) Hence the ratio is always $3$.

Olympiad / Challenge

1. 1
   Suppose $ab - 1 \mid a^2 + b^2$ with $ab > 1$, and let $k = \frac{a^2 + b^2}{ab - 1}$ be a fixed positive integer, so $a^2 + b^2 = k(ab - 1)$, i.e. $a^2 - (kb)\,a + (b^2 + k) = 0$. Among all positive-integer solutions of this equation for this $k$, choose one minimizing $a + b$, and label it so that $a \ge b \ge 1$.
2. 2
   Treat $a^2 - (kb)\,a + (b^2 + k) = 0$ as a quadratic in $a$ with the known integer root $a$. The second root is $a' = kb - a = \frac{b^2 + k}{a}$. The first form shows $a'$ is an integer; the second shows $a' > 0$ (as $b^2 + k > 0$), so $a'$ is a positive integer and $(a', b)$ is again a positive-integer solution with the same $k$.
3. 3
   Since $a + b$ was minimal over all solutions, $(a', b)$ cannot be smaller: $a' + b \ge a + b$, so $a' \ge a$. Equivalently $a\,a' = b^2 + k \ge a^2$, i.e. $k \ge a^2 - b^2$.
4. 4
   Now suppose for contradiction the minimal pair had $a > b \ge 2$. We compare $k$ with $a^2 - b^2$ directly: $(a^2 - b^2)(ab - 1) - (a^2 + b^2) = a\big(b(a-b)(a+b) - 2a\big)$, and since $a > b \ge 2$ we have $a - b \ge 1$ and $b \ge 2$, so $b(a-b)(a+b) \ge 2(a+b) > 2a$, making this positive. Hence $(a^2 - b^2)(ab - 1) > a^2 + b^2 = k(ab - 1)$, so $k < a^2 - b^2$ — contradicting $k \ge a^2 - b^2$ from the previous step. So the minimal pair cannot have $a > b \ge 2$.
5. 5
   The remaining possibilities for the minimal pair are $a = b$ or $b = 1$. If $a = b$ then $2a^2 = k(a^2 - 1)$, so $k = 2 + \frac{2}{a^2 - 1}$, which is an integer only if $(a^2 - 1) \mid 2$, i.e. $a^2 \in \{2, 3\}$ — impossible for an integer $a \ge 2$ (and $a = b = 1$ gives $ab - 1 = 0$). So the minimal pair has $b = 1$.
6. 6
   With $b = 1$ the equation is $a^2 + 1 = k(a - 1)$, so $(a - 1) \mid a^2 + 1 = (a - 1)(a + 1) + 2$, forcing $(a - 1) \mid 2$ and hence $a \in \{2, 3\}$. The base pairs are $(2, 1)$ and $(3, 1)$, giving $\frac{2^2 + 1}{2 \cdot 1 - 1} = 5$ and $\frac{3^2 + 1}{3 \cdot 1 - 1} = 5$. Every solution descends to one of these (and their reflections $(1,2),(1,3)$), so $k = 5$ for every solution.