AoPS - Problem

Let $ a \le b \le c$ be the sides, $ a = 2007 = 3^2 \cdot 223,$ and $ \angle C = 2 \angle A.$ By the sine and cosine theorems, $ \frac {c}{a} = \frac {\sin 2A}{\sin A} = 2 \cos A$ $ a^2 = b^2 + c^2 - \frac {bc^2}{a},\ \ c^2 (b - a) = a(b^2 - a^2) = a(b + a)(b - a)$ If $ b = a$ and $ \angle C = 2 \angle A = 2 \angle B,$ the $ \triangle ABC$ is isosceles right and the largest side $ c$ is not an integer. Thus $ a < b \le c,$ $ c^2 = a(a + b).$ But $ c^2 = a(a + b) = 3^2 \cdot 223 (a + b)$ implies $ c = 3 \cdot 223 \cdot n,\ n$ is an integer. Then $ 3^2 \cdot 223^2 \cdot n^2 = 3^2 \cdot 223 (3^2 \cdot 223 + b),$ $ b = 223 (n^2 - 3^2).$ Thus $ (a, b, c) = 223 \cdot (3^2, n^2 - 3^2, 3n)$ From $ c > a,$ we have $ 3n > 3^2,\ n \ge 4,$ and from $ \cos A = \frac {c}{2a} = \frac {3n}{2 \cdot 3^2} = \frac {n}{6},$ we have $ n \le 5.$ It follows that $ n^2 - 3^2 = 7$ or $ 16.$ The 1st possibility is excluded by $ b > a,\ n^2 - 3^2 > 9,$ the other one by $ b \le c,\ n^2 - 3^2 \le 3n \le 15.$