The side lengths of $\triangle ABC$ are integers with no common factor greater than $1$.  Given that $\angle B  = 2 \angle C$ and $AB < 600$, compute the sum of all possible values of $AB$.

[i]Proposed by Eugene Chen[/i]
1. **Using the Angle Bisector Theorem:**
   Given that $\angle B = 2 \angle C$, we can use the angle bisector theorem. Let the angle bisector of $\angle B$ intersect $AC$ at $D$. According to the angle bisector theorem, the ratio of the segments created by the bisector on $AC$ is equal to the ratio of the other two sides of the triangle:
   \[
   \frac{AD}{DC} = \frac{AB}{BC} = \frac{c}{a}
   \]
   Let $AD = x$ and $DC = y$. Then:
   \[
   \frac{x}{y} = \frac{c}{a} \implies x = \frac{c}{a} y
   \]
   Since $AD + DC = AC = b$, we have:
   \[
   x + y = b \implies \frac{c}{a} y + y = b \implies y \left( \frac{c}{a} + 1 \right) = b \implies y = \frac{ab}{a+c}
   \]
   Therefore:
   \[
   x = \frac{c}{a} y = \frac{c}{a} \cdot \frac{ab}{a+c} = \frac{c b}{a+c}
   \]
   So, $AD = \frac{c b}{a+c}$.

2. **Using the Law of Sines:**
   Since $\angle B = 2 \angle C$, we can use the Law of Sines in $\triangle ABC$:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
   \]
   Let $\angle C = \theta$. Then $\angle B = 2\theta$ and $\angle A = 180^\circ - 3\theta$. Using the Law of Sines:
   \[
   \frac{a}{\sin(180^\circ - 3\theta)} = \frac{b}{\sin 2\theta} = \frac{c}{\sin \theta}
   \]
   Since $\sin(180^\circ - x) = \sin x$, we have:
   \[
   \frac{a}{\sin 3\theta} = \frac{b}{2 \sin \theta \cos \theta} = \frac{c}{\sin \theta}
   \]
   Using the triple angle formula $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$, we get:
   \[
   \frac{a}{3 \sin \theta - 4 \sin^3 \theta} = \frac{b}{2 \sin \theta \cos \theta} = \frac{c}{\sin \theta}
   \]
   Simplifying, we find:
   \[
   a = \frac{c (3 \sin \theta - 4 \sin^3 \theta)}{\sin \theta} = c (3 - 4 \sin^2 \theta)
   \]
   Since $\sin^2 \theta = \frac{c^2}{b^2}$, we have:
   \[
   a = c \left( 3 - 4 \frac{c^2}{b^2} \right) = \frac{3c b^2 - 4c^3}{b^2}
   \]

3. **Simplifying the Expression:**
   Given that $a = \frac{b^2}{c} - c$, we can rewrite the equation:
   \[
   c(a + c) = b^2
   \]
   This implies that $c$ must be a perfect square. Let $c = k^2$. Then:
   \[
   a = \frac{b^2}{k^2} - k^2
   \]
   For $a$ to be an integer, $b$ must be a multiple of $k^2$. Let $b = k^2 m$. Then:
   \[
   a = \frac{(k^2 m)^2}{k^2} - k^2 = k^2 (m^2 - 1)
   \]

4. **Finding Possible Values of $AB$:**
   Since $AB < 600$, we need to find all possible values of $b = k^2 m$ such that $b < 600$. We also need to ensure that $a$, $b$, and $c$ are coprime. The possible values of $k$ are integers such that $k^2 < 600$. The largest such $k$ is $24$.

   We sum the values of $b = k^2 m$ for $k = 2, 3, \ldots, 24$ and $m$ such that $k^2 m < 600$.

The final answer is $\boxed{None}$.