# How many right triangles, whose sides are all positive whole numbers, have the property that the area is numerically equal to perimeter?

Jul 28, 2018

Let $a , b \mathmr{and} c$ are three sides of a right triangle where $c$ is the hypotenuse , This means $c > a \mathmr{and} c > b$.

As per given condition of the problem $a , b \mathmr{and} c$ are all positive whole numbers ,
and

$\frac{1}{2} a b = a + b + c \ldots . . \left[1\right]$

Again by Pythagorean theorem

${c}^{2} = {a}^{2} + {b}^{2.} \ldots \ldots . \left[2\right]$

Combining [1] and [2] we get

$\implies {\left(\frac{1}{2} a b - a - b\right)}^{2} = {a}^{2} + {b}^{2}$

$\implies \frac{1}{4} {a}^{2} {b}^{2} + {a}^{2} + {b}^{2} - 2 \cdot \frac{1}{2} {a}^{2} b - 2 \cdot \frac{1}{2} a {b}^{2} + 2 a b = {a}^{2} + {b}^{2}$

$\implies \frac{1}{4} {a}^{2} {b}^{2} - {a}^{2} b - a {b}^{2} + 2 a b = 0$

$\implies \frac{1}{4} a b - a - b + 2 = 0.$ as $a b \ne 0$

$\implies \frac{1}{2} a b - 2 a - 2 b + 4 = 0. \ldots \ldots . \left[3\right]$

Combining {1} and {3] we get

$2 a + 2 b - 4 = a + b + c .$

$\implies c = a + b - 4. \ldots . . \left[4\right]$

Combining [1] and [4] we get

$2 a + 2 b - 4 = \frac{1}{2} a b$

$\implies a + b = 2 + \frac{1}{4} a b \ldots \ldots \left[5\right]$

LHS of this relation is an integer. To satisfy this both $a \mathmr{and} b$ should be even or any of $a \mathmr{and} b$ is a multiple of $4$. So minimum value of $a \mathmr{and} b = 4$. If minimum value of $b = 4$ then minimum value of $a$ will be $3$ and $c = 5$ But it does not satisfy equation [1]
For integer values of $a , b \mathmr{and} c$ satisfying these conditions and equation {2] we get the following when $b = 8$,the next integral multiple of $4$

$a = 6 , b = 8 \mathmr{and} c = 10$

For $b$ taking other higher multiple of $4$ as 12,16,20..etc the relation (5) is not satisfied.