
3. Find all integers $a, b, c$ such that

$$
a^{2}=b c+4, \quad b^{2}=c a+4
$$


Solution: Suppose $a=b$. Then we get one equation: $a^{2}=a c+4$. This reduces to $a(a-c)=4$. Therefore $a=1, a-c=4 ; a=-1, a-c=-4 ; a=4, a-c=1$; $a=-4, a-c=-1 ; a=2, a-c=2 ; a=-2, a-c=-2$. Thus we get $(a, b, c)=(1,1,-3)$, $(-1,-1,3),(4,4,3),(-4,-4,-3) ;(2,2,0),(-2,-2,0)$.

If $a \neq b$, subtracting the second relation from the first we get

$$
a^{2}-b^{2}=c(b-a)
$$

This gives $a+b=-c$. Substituting this in the first equation, we get

$$
a^{2}=b(-a-b)+4
$$

Thus $a^{2}+b^{2}+a b=4$. Multiplication by 2 gives

$$
(a+b)^{2}+a^{2}+b^{2}=8
$$

Thus $(a, b)=(2,-2),(-2,2),(2,0),(-2,0),(0,2),(0,-2)$. We get respectively $c=$ $0,0,-2,2,-2,2$. Thus we get the triples:

$$
\begin{array}{r}
(a, b, c)=(1,1,-3),(-1,-1,3),(4,4,3),(-4,-4,-3),(2,2,0),(-2,-2,0) \\
(2,-2,0),(-2,2,0),(2,0,-2),(-2,0,2),(0,2,-2),(0,-2,2)
\end{array}
$$
