Find the sides of the triangle.

Question

The triangle with sides $8-15-13$ has a $60^{\circ}$ angle. The triangle with sides $11-35-31$ also has a $60^{\circ}$ angle. Find a triangle $x-y-403$ where $x$ and $y$ are relatively prime positive integers and the angle opposite the side of length $403$ is $60^{\circ}$ angle.

I have been trying this problem but doesn't get the solution. I am unable to find how I can use the given info to solve this.

Is the given information sufficient to solve the problem?

If yes, can anybody please tell me how I can use them.

Thanks in advance.

yeah, that you will get from law of cosines. But where is the given information used? And how to solve it further? — User, Jun 23 '16 at 04:39
We have a Diophantine equation. It has $x=80, y=437$ as a solution, maybe others. — André Nicolas, Jun 23 '16 at 04:58

score 2 · Answer 1 · answered Jun 23 '16 at 15:31

We are looking for the integer solutions of $ \frac{x^2+y^2-403^2}{2xy}=\frac{1}{2}$ or $$ x^2-xy+y^2 = 403^2 = 13^2\cdot 31^2.\tag{1} $$ It is crucial to point out that the LHS of $(1)$ is the norm on the Euclidean domain $\mathbb{Z}[\omega]$ (the ring of Eisenstein integers) that is a unique factorization domain. $31$ splits as $(6+\omega)(6+\omega^2)$ and $13$ splits as $(4+\omega)(4+\omega^2)$, hence the solutions of $(1)$ over the positive integers, with $x\leq y$, are given by:

$$(x,y)\in\color{red}{\left\{( 403 , 403 ), ( 80 , 437 ), ( 357 , 437 ), ( 115 , 448 ) , ( 333 , 448 ) ,\\ ( 143 , 455 ) , ( 312 , 455 ) , ( 217 , 465 ) , ( 248 , 465 ) \right\}}\tag{2} $$

Find the sides of the triangle.

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