
3. The present ages in years of two brothers $A$ and $B$, and their father $C$ are three distinct positive integers $a, b$, and $c$ respectively. Suppose $\frac{b-1}{a-1}$ and $\frac{b+1}{a+1}$ are two consecutive integers, and $\frac{c-1}{b-1}$ and $\frac{c+1}{b+1}$ are two consecutive integers. If $a+b+c \leq 150$ determine $a, b$ and $c$.


Solution: We have

$$
\frac{b-1}{a-1}=l, \quad \frac{b+1}{a+1}=l-1, \quad \frac{c-1}{b-1}=m, \quad \frac{c+1}{b+1}=m-1
$$

(If we take $a \leq b$, we see that $\frac{b-1}{a-1} \geq \frac{b+1}{a+1}$.)

Consider the first two relations: $b-1=l(a-1), b+1=(l-1)(a+1)$. Solving for $a$, we get $a=2 l-3$ and hence $b=2 l^{2}-4 l+1$. Using the second set of relations, we obtain $b=2 m-3$ and $c=2 m^{2}-4 m+1$. Thius we have $2 m-3=2 l^{2}-4 l+1$ or $m=(l-1)^{2}+1$. Obviously $l>1$. If $l=2$, we get $a=1$ which forces $b=1$ and $c=1$, which is impossible. If $l=3$, we get $a=3, b=7$ and $c=31$. If $l \geq 4$, then $m \geq 10$ and $c \geq 161$. But then $a+b+c>150$. Thus the only choice is $a=3, b=7$ and $c=31$.
