# Understand the problem

Let $f$ be a polynomial with integer coefficients. Define $a_1 = f(0)~,~a_2 = f(a_1) = f(f(0))~,$ and $~a_n = f(a_{n-1})$ for $n \geqslant 3$.

If there exists a natural number $k \geqslant 3$ such that $a_k = 0$, then prove that either $a_1=0$ or $a_2=0$.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 7.

##### Topic

Polynominals (Algebra)

8 out of 10

# Start with hints

Do you really need a hint? Try it first!

Do you know this lemma ,

Lemma: If $p, q \in \mathbb{Z}$ and $p \neq q$, then $p - q \mid f(p) - f(q)$ .

To prove this,

let $f(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_0$. Then $f(p) - f(q) = a_n(p^n - q^n) + a_{n-1}(p^{n-1} - q^{n-1}) + a_{n-2}(p^{n-2} - q^{n-2}) + \cdots + (p - q).$Each bracket is divisible by $p - q$, proving the statement.

We use the fact that the sequence $a_1, a_2, a_3, \cdots$ consists of only integers.
We’ll first prove that we cannot have three distinct integers $p$, $q$, and $r$ such that $f(p) = q$, $f(q) = r$, and $f(r) = p$ (In other words, the variables cannot come in a cycle of 3). Assume that there does exist such numbers. Then we should have $p - q \mid f(p) - f(q) = q - r$, which means $\mid p - q \mid \le \mid q - r \mid$ . Similarly we can get $\mid p - q \mid \le \mid q - r \mid \le \mid r - p\mid \le \mid p - q \mid$ , 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 $f$ can only come in cycles of most two.

We realize that since $a_{k+1} = f(0) = a_1$,

we have a cycle $a_1, a_2, a_3, \cdots, a_k$. Since the minimal cycle has length at most 2, one of $a_1$ or $a_2$ must be equal to 0, and we are done.

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