Prove that $a$ cannot be a prime The sides of a triangle are of length $a,b,c$ where $a,b,c$ are integers and $a>b$, angle opposite to $c$ is $60$ degrees. Prove that $a$ Cannot be a prime

By the Law of Cosines, we have
$$
c^2=a^2+b^2-2ab\cos(\angle ACB)=a^2+b^2-ab\implies(c+b)(c-b)=a(a-b).
$$
Now, suppose $a$ is prime. Then, either $a|c+b$ or $a|c-b$. But by the triangle inequality, $a>c-b$ so we must have 
$$
a|c+b. 
$$
Next, because the angle opposite $c$ is $60$ degrees, either of the remaining angles is at least as large as $60$. Thus, by the Law of Sines, we can infer that $c\leq\max\{a,b\}=a$. This, together with $b<a$, implies that $c+b<2a$. But $a|c+b$ so we must have $a=c+b$, violating the triangle inequality and giving us the desired contradiction.
