Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$

Question

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 $a, b, c$.

Answer 1

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.

Answer 2

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.

Answer 3

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!.