IRL_ABooklet

Let $a,b,c$ be the side lengths of a right-angled triangle with hypotenuse $c$, i.e. $c^2=a^2+b^2$. The area of this triangle is equal to $ab/2$ and the perimeter is $a+b+c$. Area and perimeter agree exactly when $ab/2=a+b+c$. Squaring both sides and using Pythagoras, this becomes $$a^2+b^2={\left(\frac{ab}{2}-(a+b)\right)}^2=\frac{a^2{b}^2}{4}+(a+b{)}^2-ab(a+b).$$ We simplify this to $$0=\frac{a^2{b}^2}{4}+2ab-ab(a+b)=ab\left(\frac{ab}{4}+2-(a+b)\right).$$ Because $a$ and $b$ are positive, we obtain $ab+8-4a-4b=0$, which is equivalent to $(a-4)(b-4)=8.$ We may assume, w.l.o.g., that $1\le a\le b$. If $b\le 4$, then both factors, $a-4$ and $b-4$, are between 0 and $-3$. The product of two such integers is never equal to 8. Hence, $b\ge 5$, which implies that $b-4$ is positive. Therefore, $a-4$ must be positive as well, hence $5\le a\le b$. The only factorisations of 8 in positive integers are $8=1\cdot 8$ and $8=2\cdot 4$. We obtain $(a,b)=(5,12)$ in the first case, and $(a,b)=(6,8)$ in the second case. These lead to the right-angled triangles with side lengths 5, 12, 13 and 6, 8, 10. A straightforward check reveals that these are indeed solutions to this problem.