## Pythagorean Triples With Area Equals Perimeter

For which Pythagorean Triples is the area of the associated right angled triangle equal to it's perimeter?
All Pythagorean triples are generated by
$(m^2-n^2,2mn,m^2+n^2$
.
The perimeter of the triangle is
$m^2-n^2+2mn+m^2+n^2=2mn+2m^2=2m(n+m)$
.
The area of the triangle is
$\frac{1}{2} (2mn)(m^2-n^2)=mn(m^2-n^2)=mn(m+n)(m-n)$
.
Equating these gives
$2m(m+n)=mn(m+n)(m-n)$

$2m(m+n)-mn(m+n)(m-n)=0$

$m(m+n)(n^2-mn+2)=0$

We must have
$n^2-mn+2=0 \rightarrow n=\frac{--m \pm \sqrt{m^2-8}}{2}$
.
$m^2-8$
must be a square number and the only possibility is
$m=3 \rightarrow \sqrt{3^2-8}=1$
then
$n= \frac{--3 \pm 1}{2}=2, \; 1$
. The only possible triangles are
$(3^2+2^2,2 \times 3 \times 2,3^2+2^2)=(5,12,13)$

$(3^2-1^2,2 \times 3 \times 1,3^2+1^2)=(8,6,10)$