503. Find a triangle with sides $a, b$ and $c$ and area $S$ expressed by four consecutive integers.

## Problem from "L'Education mathématique".
503. Let the sides of the desired triangle be $x-1, x, x+1$, and the area $x+2$, then

$$
\begin{gathered}
\sqrt{3 x(x+2)(x-2) x}=4(x+2) \\
3 x^{2}\left(x^{2}-4\right)=16(x+2)^{2} \\
3 x^{2}(x-2)=16(x+2) \\
3 x^{3}-6 x^{2}-16 x-32=0 \\
3 x^{3}-12 x^{2}+6 x^{2}-24 x+8 x-32=0 \\
3 x^{2}(x-4)+6 x(x-4)+8(x-4)=0 \\
(x-4)\left(3 x^{2}+6 x+8\right)=0 \\
x_{1}=4 ; x_{2,3}=\frac{-3 \pm \sqrt{-15}}{3}
\end{gathered}
$$

Only the value $x_{1}$ satisfies, hence the sides of the triangle are $3,4,5$, and the area is 6. Clearly, the triangle is a right triangle.