The sides of a right-angled triangle are integers. Twice the measure of the area of the triangle is equal to three times the measure of the perimeter. What are the lengths of the sides of the triangle?
I. solution: Let the legs of the triangle be denoted by $a$ and $b$, and the hypotenuse by $c$. We can assume that $a\varrho=3$, so $x-3, y-3$ are positive and $a\varrho < x, y$. This leads us to a Diophantine equation. From this, we obtain as many solutions as there are ways to factorize $2 \varrho^{2}$ into the product of two positive integers, since $\varrho, x-\varrho, y-\varrho$ together with $x, y, a=x+\varrho, b=y+\varrho$ and $c=x+y$ are also integers.

It is interesting that for every value of $\varrho$, there is a solution where the sides are in the ratio $3: 4: 5$. This is due to the factorization of $2 \varrho^{2}$ as $\varrho \cdot 2 \varrho$: from $x-\varrho=\varrho, y-\varrho=2 \varrho$ we get $x=2 \varrho, y=3 \varrho$ and $a=3 \varrho, b=4 \varrho, c=5 \varrho$.

III. solution: It is known that the triples of integers $a, b, c$ that satisfy equation (1) are called Pythagorean triples. Every multiple of $a, b, c$ by a positive integer $\lambda$ is also a Pythagorean triple. If $a$ and $b$ are coprime, then $c$ and $a$, and $c$ and $b$ are also coprime, in which case $a, b, c$ is called a primitive triple. It is also known that every primitive triple can be expressed with two parameters $u, v$ as follows:

$$
a=u \cdot v, \quad b=\frac{u^{2}-v^{2}}{2}, \quad c=\frac{u^{2}+v^{2}}{2}
$$

where $u$ and $v$ are positive integers, odd and coprime. With these, every Pythagorean triple can be written as:

$$
a=\lambda \cdot u \cdot v, \quad b=\lambda \cdot \frac{u^{2}-v^{2}}{2}, \quad c=\lambda \cdot \frac{u^{2}+v^{2}}{2}
$$

since every Pythagorean triple is a multiple of some primitive triple.

Our task now is to select from (7) the triples that satisfy (2). (One of $av$ is even, the other is odd. See, for example: Rademacher-Toeplitz: On Numbers and Figures, Middle School Special Interest Booklets, Tankönyvkiadó, 1954, p. 84.

Among the factors on the left side, $v$ is odd, $u-v$ is even, and $\lambda$ is any positive integer. On the other hand, $6=2 \cdot 3=1 \cdot 2 \cdot 3$, so the following possibilities exist for the correspondence of the factors:

$$
\begin{aligned} &\(v=\overbrace{2,}^{1,} 6 \\ &, u-v=\quad 3 \\ & 2\end{aligned}
$$\),

then
$\lambda=$
3,
1,
1,

and

$$
u=\quad 3, \quad 7, \quad 5
$$

therefore, according to (7),

| $a=$ | 9, | 7, | 15, |
| ---: | ---: | ---: | ---: |
| $b=$ | 12, | 24, | 8, |
| $c=$ | 15, | 25, | 17, |

We arrive again at the triangles we obtained earlier. Furthermore, we see from the value $\lambda=1$ that the second and third solutions are primitive triples. Indeed, the numbers $7,24,25$ and $15,8,17$ are pairwise coprime, while the first triple is three times the primitive triple $3,4,5$.


[^0]:    ${ }^{1}$ Similar, more well-known expressions are the following:

    $$
    a=u^{2}-v^{2}, \quad b=2 u v, \quad c=u^{2}+v^{2}
    $$