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  
The perimeter of the triangle is  
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



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

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

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