The proof I wrote above is logically incorrect. I have got the right way to prove it now. Administrator may close the thread.
I found the following exercise in Chartrand's book at the end of chapter 3, and attempted to prove it, but I am not very satisfied because it does not seem legitimate. I have not officially learned the mathematical proofs so kindly please help me.
Exercise 3.45:
Prove that if and are two positive integers, then
The following is my weak attempt:
Let , and
The algebraic reduction of
We can restate the question as and
Proof:
By trichotomy law, we have three cases, namely and
Case 1: Assume that Since
Case 2: Assume that . Since the square of any integer is a positive integer then and . It follows that
Since Case 3 is similar to case 2, we will not repeat. Q.E.D