Let $a$ and $b$ be positive integers such that φ(ab + 1) divides a^2 + b^2 + 1. Prove that $a$ and $b$ are Fibonacci numbers.

Step 1 (Define the Constant Ratio): Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2+1$. This means there exists a positive integer $k$ such that:

Step 2 (Vieta Descent / Infinite Descent): Fix $k$ and assume there exists a solution $(a,b)$ with $a\ge b\ge 1$. Consider the quadratic equation in $x$:

By Vieta's formulas, if $a$ is a root, the other root $x_2$ satisfies:

This implies $x_2=kb-a$ is an integer. Using standard inequalities on the roots, if $b>1$, we must have $0<x_2<a$, which yields a strictly smaller solution $(b,x_2)$ with $b\ge x_2$. This descent must terminate at $b=1$.

Step 3 (Base Case and Fibonacci Relation): When $b=1$, the equation becomes:

For $a\ge 1$, the only integer solution is $k=5$ (which yields $a=3$ or $a=2$). If $k=5$, the relation becomes $a^2-5ab+b^2-4=0$, which is the characteristic equation for consecutive Fibonacci numbers of odd index ($F_{2m-1},F_{2m+1}$). Thus, $a$ and $b$ are consecutive Fibonacci numbers. $■$