Understand the problem
Let be a polynomial with integer coefficients. Define and for .
If there exists a natural number such that , then prove that either or .
Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 7.
8 out of 10
Start with hints
Do you really need a hint? Try it first!
Do you know this lemma ,
Lemma: If and , then .
To prove this,
let . ThenEach bracket is divisible by , proving the statement.
We use the fact that the sequence consists of only integers.
We’ll first prove that we cannot have three distinct integers , , and such that , , and (In other words, the variables cannot come in a cycle of 3). Assume that there does exist such numbers. Then we should have , which means . Similarly we can get , which implies equality. Ultimately, it leads to two equal variables, contradiction. In a similar manner we can prove that these variables cannot come in cycles of more than 3.
Therefore, we conclude that the variables of can only come in cycles of most two.
We realize that since ,
we have a cycle . Since the minimal cycle has length at most 2, one of or must be equal to 0, and we are done.
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