Find all integer pairs $(a,b)$ such that $a^2+ab+b^2=219^2$, without technology.

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Find all integer pairs $(a,b)$ such that $a^2+ab+b^2=219^2$, without technology.

My colleague showed me this question. He said it is from the AIME (American Invitational Mathematics Examination).

I regarded the equation as a quadratic in $b$, so $$b=\frac{-a\pm\sqrt{438^2-3a^2}}{2}$$ so $438^2-3a^2$ must be a square number. Without technology, I have not found a nice way to find all possible values of $a$.

EDIT: I don't know if this helps, but if we ignore the bit about "without technology", then using excel we can find that the solutions are:

$\pm(0, 219), \pm(51, 189), \pm(51, -240), \pm(189,-240), (219, -219)$ and their inverses.

EDIT2: My colleague, to my annoyance, just told me that the whole question was this: A triangle has side lengths $a, b, 219$ where $a$ and $b$ are integers. The angle opposite the side of length $219$ is $120°$. Find $a$ and $b$.

I apologize for posting an incomplete, possibly unsolvable, question and thus wasting readers' time. In future, when posting questions that I get from other people, I will be more careful to avoid the X-Y problem.

Having said that, I am still interested in whether the original version of the question is solvable.