147. Diophantine system. Solve the following system of equations in positive integers:

$$
\left\{\begin{aligned}
a^{3}-b^{3}-c^{3} & =3 a b c \\
a^{2} & =2(b+c)
\end{aligned}\right.
$$
147. The first equation of the system can be written as

$$
(a-b-c)\left[a^{2}+(b-c)^{2}+a b+b c+c a\right]=0
$$

Since the second factor cannot be zero for positive \(a, b, c\), we obtain

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

Therefore, the only solution to our system in positive integers is \(a=2, b=c=1\).

[E. W. Marchand, A. M. M., 65, 43 (January 1958).]