Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse. 
I came up with 2 equations which are:
$\dfrac{ab}2 = 2(a+b+c)\;\;$  and  $\;\;a^2+b^2=c^2$
but I'm not sure how to proceed and solve for $a, b, c$.

Rewrite the first equation as $c = \frac{ab}{4} - a - b$. Square it to get $$c^2 = a^2 + b^2 + \frac{a^2b^2}{16} - \frac{a^2b}{2} - \frac{ab^2}{2} + 2ab$$
Now using the other equation, we see that
$$\frac{a^2b^2}{16} - \frac{a^2b}{2} - \frac{ab^2}{2} + 2ab = 0$$
Since $a,b > 0$ divide by $ab$ and multiply by $16$ to get
$$ ab - 8a - 8b + 32 = 0$$
Use Simon's Favorite Factoring Trick to get $(a-8)(b-8) = 32$.
Now note that $a$ and $b$ are integers, so $(a-8)$ and $(b-8)$ must be factors of $32$. But factoring $32$ into anything except for $\{1,32\}$ gives you two even numbers - these can't be the legs of a primitive Pythagorean triple. Thus, we must have $a=9,b=40$, giving us the $(9,40,41)$ triangle.
