2013中国女子数学奥林匹克试题及其解答(8)

Now by Cauchy-Schwarz inequality we have

and if

, we have

and we need to show that

. Assume

. Then we have

which means that

4.解（crazyfehmy）

Let

which is a contradiction. So, we are done. be called a nice pair if satisfies the conditions stated in the problem.

Firstly, we shall prove a lemma:

If

Proof:

Let

or and

divides for some integers

such

that

and and

is a nice pair then for all integers and . . Then we can find another

pair

. (The proof is easy) Then consider the system

and

Remainder Theorem this system has a solution an element

of and . By the Chinese such that both and is is not a nice pair

because

which means

that

is divisible by

Now, we shall show that

if

solution for all integers or

then . has a such

that . For the proof assume there exists an integer

has no solution in integers. Then it is easy to see

that

has also no solution for all integers which are not divisible