## Task 18/84

We are looking for all right-angled triangles with integer side lengths, where the numerical values of the area and perimeter are equal.
According to the problem statement, all Pythagorean number triples $(a ; b ; c)$ are sought for which

$$
\frac{a b}{2}=a+b+c
$$

(swap of $a$ and $b$ is irrelevant). It is well known that all Pythagorean triples $(a ; b ; c)$ can be obtained through the approach

$$
a=k\left(m^{2}-n^{2}\right) \quad ; \quad b=2 k m n \quad ; \quad k\left(m^{2}+n^{2}\right)
$$

with $k ; m \in \mathbb{N}, k ; m ; n \neq 0$. Substituting this, we get

$$
\frac{k\left(m^{2}-n^{2}\right) \cdot 2 k m n}{2}=k\left(m^{2}-n^{2}\right)+2 k m n+k\left(m^{2}+n^{2}\right) \rightarrow k^{2}(m-n)(m+n) m n=2 k m(m+n)
$$

or, due to $(m+n) ; k ; m ; m \neq 0$

$$
k(m-n) n=2 \rightarrow m=n+\frac{2}{k n}
$$

For $m \in \mathbb{N}$ to hold, $k n=1$ or $k n=2$ must be true, so $k_{1}=1, n_{1}=1, k_{2}=1, n_{2}=2$ or $k_{3}=2, n_{3}=1$. Then $m_{1}=m_{2}=3, m_{3}=2$ and it follows that

$$
\begin{array}{ccc}
a_{1}=8 & b_{1}=6 & c_{1}=10 \\
a_{2}=5 & b_{2}=12 & c_{2}=13 \\
a_{3}=6 & b_{3}=8 & c_{3}=10
\end{array}
$$

One can see that the third solution (apart from the irrelevant swap of $a$ and $b$) coincides with the first. The problem thus has exactly two essentially different solutions, as confirmed by the verification.