Calculate the three sides of a certain triangle, if we know that the measures of the sides and the measure of the area are integers and form an arithmetic progression in the said order.
Let $x$ be the measure of the middle side of a triangle and $y$ the common difference of an arithmetic progression, the sides and the area, respectively, are as follows:

$$
x-y, x, x+y, x+2 y
$$

The area expressed by the sides is the following:

$$
\sqrt{\frac{3 x^{2}(x+2 y)(x-2 y)}{16}}
$$

and according to the assumption,

$$
x+2 y=\sqrt{\frac{3 x^{2}(x+2 y)(x-2 y)}{16}}
$$

or

$$
(x+2 y)^{2}=\frac{3 x^{2}(x+2 y)(x-2 y)}{16}
$$

Since $x+2 y>0$, because $x>0$ and $y>0$, both sides of the latter equation can be divided by $x+2 y$. Thus, we have:

$$
x+2 y=\frac{3 x^{2}(x-2 y)}{16}
$$

from which

$$
y=\frac{3 x^{3}-16 x}{2\left(3 x^{2}+16\right)}
$$

For $y$ to be an integer, the numerator of its fractional form must be even, since the denominator is also even. But the numerator can only be even if $x$ itself is even. Therefore, we can write:

$$
x=2 z
$$

Substituting this value into 1), we get the following equation:

$$
y=\frac{3 z^{3}-4 z}{3 z^{2}+4}
$$

or

$$
y=z-\frac{8 z}{3 z^{2}+4}
$$

Since $y$ must be positive, from 2) we get:

$$
3 z^{3}-4 z>0
$$

or

$$
3 z^{2}-4 z>0
$$

and thus

$$
z>1
$$

For $y$ to be an integer, $\frac{8 z}{3 z^{2}+4}$ must be an integer, which is only possible if

$$
8 z \geq 3 z^{2}+4
$$

or

$$
z(8-3 z) \geq 4
$$

But at the same time

$$
3 z<8
$$

or

$$
z<3
$$

However, the conditions 4) and 5) are only satisfied simultaneously by the equation $z=2$. From this it follows that:

$$
\begin{aligned}
& x=4 \\
& y=1
\end{aligned}
$$

and the sides and area of the sought triangle are, respectively: