1993
Problem - 3628
Let $a, b, c, p$ be real numbers, with $a, b, c$ not all equal, such that $$a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=p$$ Determine all possible values of $p$ and prove $abc+p=0$.
Because $ac+1 = ap$ and $bc+1=cp$, it must be true that
$$ap^2=cap+p=a(bc+1)+p=abc+a+p\implies a(p^2-1)=abc+p$$
Similarly, we must have $b(p^2-1)=abc+p$ and $c(p^2-1)=abc+p$.
Because $a, b, c$ are not all equal, these three relations can hold if and only if $p^2-1=0\implies p=\boxed{\pm 1}$.
Accordingly, their right sides $abc+p$ must equal 0.