## PROBLEM 30. SYSTEM OF DIOPHANTINE EQUATIONS

Solve the following system of equations in natural numbers:

$$
\begin{aligned}
a^{3}-b^{3}-c^{3} & =3 a b c \\
a^{2} & =2(b+c)
\end{aligned}
$$
Solution. Since $3 a b c$ is a positive number, then $a^{3}$ must be greater than the sum of $b^{3}$ and $c^{3}$; thus, 52
we have

$$
b<a \quad \text { and } \quad c<a
$$

Adding these inequalities, we get that $b+c<2 a$, therefore, $2(b+c)<4 a$. From the second equation, we get that

$$
a^{2}<4 a \quad \text { and } \quad a<4
$$

But the second equation also shows that $a$ is an even number. Thus, a must be equal to 2, and the smaller numbers $b$ and $c$ must be equal to 1.