Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Prove that the integer $\frac{a^2 + b^2}{ab + 1}$is a perfect square. --- Let $k$ be the integer such that $k = \frac{a^2 + b^2}{ab + 1}$. Our objective is to prove that $k$ is a perfect square. We will proceed by contradiction, utilizing the Well-Ordering Principle. ### Step 1: The Assumption of a Minimal Counterexample Assume, for the sake of contradiction, that there exists at least one positive integer $k$ which is _not_ a perfect square, yet can be expressed as $k = \frac{a^2 + b^2}{ab + 1}$ for some positive integers $a$ and $b$. Among all such pairs of positive integers $(a, b)$ that evaluate to this specific non-square $k$, there must exist a "minimal" pair. Let us denote this minimal pair as $(A, B)$, defined such that the sum $A + B$ is minimized. Without loss of generality, we may assume that $A \ge B > 0$. ### Step 2: Constructing the Quadratic Equation We can rearrange our fundamental equation into a quadratic form. Since $k = \frac{A^2 + B^2}{AB + 1}$, we have: $A^2 - (kB)A + (B^2 -k)=0$ Fixing $B$ and $k$ as constants, we can view this as a quadratic equation in a single variable $x$: $x^2 - (kB)x + (B^2 - k) = 0$ By definition, we already know that $x_1 = A$ is a valid root of this equation. Because this is a quadratic equation, it must possess a second root, which we will call $A_2$. ### Step 3: Proof $A_2$ is a Positive Integer By Vieta's formulas, the sum and product of the roots must satisfy: 1. $A + A_2 = kB$ 2. $A \cdot A_2 = B^2 - k$ From the sum formula, we observe that $A_2 = kB - A$. Because $k, B,$ and $A$ are all integers, it strictly follows that **$A_2$ is an integer**. #### Case 1: Suppose $A_2 = 0$ If $A_2 = 0$, substituting this into the product formula yields $A \cdot 0 = B^2 - k$, which implies $k = B^2$. This directly contradicts our initial assumption that $k$ is not a perfect square. Therefore, $A_2 \neq 0$. #### Case 2: Suppose $A_2 < 0$ If $A_2$ is a strictly negative integer, then $A_2 \le -1$. We can substitute $A_2$ into the quadratic equation: $A_2^2 - kBA_2 + B^2 - k=0$ Knowing that $k \ge 1$, $B \ge 1$, and $A_2 \le -1$, we can strictly bound the terms on the left side of the equation: $A_2^2 - kBA_2 + B^2 - k \ge 1 + k + 1 -k = 2>0$ This implies the equation has no real roots. Thus, $A_2$ cannot be negative. Since $A_2$ is an integer that is neither zero nor negative, it must be a strictly positive integer. ### Step 4: The Infinite Descent We have established that $(A_2, B)$ is a valid pair of strictly positive integers that yields the same non-square $k$. We now examine the size of this new pair. Returning to the product formula: $A \cdot A_2 = B^2 - k$ Since $k$ is a strictly positive integer, we know that $B^2 - k < B^2$. Therefore: $A \cdot A_2 < B^2$ Recall our initial assumption that $A \ge B$. Because $A \ge B$, we have $A_2 < B$. Because $A_2 < B \le A$, it follows that $A_2 + B < A + B$. ### Conclusion We have constructed a new pair of positive integers $(B, A_2)$ that produces the exact same non-square $k$, but whose sum $A_2 + B$ is strictly smaller than $A + B$. This directly contradicts our foundational assumption that $(A, B)$ was the minimal pair yielding $k$. Since assuming the existence of a non-square $k$ leads to an inescapable logical contradiction across all possible bounds, the initial assumption must be false. Therefore, for all positive integers $a$ and $b$, if $ab + 1$ divides $a^2 + b^2$, the integer quantity $\frac{a^2 + b^2}{ab + 1}$ must inevitably be a perfect square. #vieta-root-jumping #infinite-descent #proof-by-contradiction #imo