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
$$a^2 + b^2 = \left(\frac{ab}{2} - (a+b)\right)^2 = \frac{a^2b^2}{4} + (a+b)^2 - ab(a+b).$$
We simplify this to
$$0 = \frac{a^2b^2}{4} + 2ab - ab(a+b) = ab\left(\frac{ab}{4} + 2 - (a+b)\right).$$
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.