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

$$
a^{2}=b c+1, \quad b^{2}=c a+1
$$


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

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

$$
a^{2}=b(-a-b)+1
$$

Thus $a^{2}+b^{2}+a b=1$. Multiplication by 2 gives

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

Thus $(a, b)=(1,-1),(-1,1),(1,0),(-1,0),(0,1),(0,-1)$. We get respectively $c=$ $0,0,-1,1,-1,1$. Thus we get the triples:

$$
(a, b, c)=(1,1,0),(-1,-1,0),(1,-1,0),(-1,1,0),(1,0,-1),(-1,0,1),(0,1,-1),(0,-1,1)
$$
