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Problem 1118

AIME late

Number theory Difficulty 5.0 Prove it Austrian Mathematical Olympiad · Austria

Let $a$ and $b$ be positive integers and $c$ be a positive real number satisfying
$\frac{a+1}{b+c} = \frac{b}{a}.$
Prove that $c \ge 1$ holds.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

$\begin{aligned} a^2 + a &= b^2 + bc \\ 4a^2 + 4a + 1 &= 4b^2 + 4bc + 1 \\ (2a + 1)^2 &= 4b^2 + 4bc + 1. \end{aligned}$
Assume to the contrary that $c < 1$ holds. This yields
$(2b)^2 = 4b^2 < (2a + 1)^2 = 4b^2 + 4bc + 1 < 4b^2 + 4b + 1 = (2b + 1)^2.$
This is a contradiction as the square of an integer cannot lie strictly between two consecutive square numbers. Therefore, $c \ge 1$ holds (for instance, $a = b$ yields $c = 1$ and therefore there is a solution of the equation with $c \ge 1$ ).

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