Irish Mathematical Olympiad

Assume for the moment that $a\le b\le c$. So $$a^2+b^2=c^2.(1)$$ The condition on the area entails $ab=2(a+b+c)$ which implies that $$4c^2=(ab{)}^2-4ab(a+b)+4(a^2+b^2+2ab).(2)$$ Equations (1) and (2) imply that $$ab-4(a+b)+8=0$$ or $$ab-4(a+b)+16=8.$$ Thus $$(a-4)(b-4)=8.(3)$$ Now, there are only two ways to factorise $8$ in positive integers. Thus, either $(a,b)=(5,12)$ or $(a,b)=(6,8)$. Thus, there are (up to permutations) two triples that satisfy the given conditions namely, $(5,12,13)$ and $(6,8,10)$. $\square$