新西兰数学奥林匹克总计3道题目, 考试时间为90分钟.
设 $$a, b, c, {aover b} + {b over c} + {c over a}, {a over c} + {c over b} + {b over a}$$ 均为正整数. 证明: $a = b = c$.
解答:
易知若 $(a. b, c) e1$, 则可对 ${aover b} + {b over c} + {c over a}$ 及 ${a over c} + {c over b} + {b over a}$ 进行约分. 因此, 不失一般性, 假设 $(a, b, c) = 1$.
事实上, 若 $(a, b, c) = 1$, 则必有 $(a, b)= 1, (b, c) = 1, (c, a) = 1$. 理由是:
设 $p$ 是一个素数, 且有 $p ot| a$, $p^n ig{|}ig{|} (b, c)$ (即 $p^n ig{|} (b, c)$ 但 $p^{n+1} otig{|} (b, c)$).
令 $b = b_0p^n$, $c = c_0p^n$, ($p ot| (b_0, c_0)$). 则有 $${aover b} + {b over c} + {c over a} = {a over b_0p^n} + {b_0 over c_0} + {c_0p^n over a} = {a^2c_0 + ab_0^2p^n + b_0c_0^2p^{2n} over ab_0c_0p^n}inmathbf{N^*}Rightarrow p^n ig{|} c_0.$$ 同理可证, $p^n ig{|} b_0$. 因此 $p^n ig{|} (b_0, c_0)$ 与 $p ot| (b_0, c_0)$ 矛盾.
即 $(a, b, c) = 1Rightarrow (a, b)= 1, (b, c) = 1, (c, a) = 1$ 得证.
若 $a, b, c$ 两两互素, 则有 $${aover b} + {b over c} + {c over a} = {a^2c + b^2a + c^2b over abc}inmathbf{N^*}Rightarrow a ig{|} c^2bRightarrow a = 1.$$ 同理可得 $b= 1$, $c = 1$. 因此 $a = b =c$.
Q.E.D.