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 $a, b, c$.

Hint: isolate for c in both equations: 
$c=\frac{ab}4  - a - b$ and $c = \sqrt{a^2+b^2} $.  Now isolate for a or b and then substitute back into your equations.
