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: formula for primitive triples is 
$a = mn; b = \frac {m^2 -n^2}2; c = \frac {m^2 + n^2}2$ for $\gcd(m,n) = 1$. And as $m^2 - n^2 $ is even $m$ and $n$ must both be odd.
So we want $\frac {mn(m^2 -n^2)}4 = 2(mn + \frac {m^2 - n^2}2 + \frac {m^2 - n^2} 2) = 2(mn + m^2)$
So $mn(m-n)(m+n) = 8m(m+n)$
So $n(m-n)=8$
For $n=1,2,4,8$ we have $m = 9,6,6,9$.  Only $n =1; m= 9$ have $\gcd(m,n) =1$ and $m^2 -n^2$ is even.
So solutions are $a = 9; b = 40; c=41$ 
====earlier answer with and an easier but not exhaustive formula for a triple ====
==== it worked out nicely, but it doesn't rule out that there aren't any others; nor did it guarentee I'd find a solution ===
Hint: Formula for primitive triples is $k, \frac {k^2 - 1}2, \frac {k^2+1}2$ where $k$ is odd.  Ex.  $3,4,5; 5,12,13; 7,24,25$ etc. 
So we need to solve $\frac {ab}2 = \frac {k (k^2 -1)}4 = 2(a+b+c) = 2(k + \frac {k^2 - 1}2 + \frac {k^2 + 1}2) = 2(k + k^2)$.
Has a very nice solution!.
