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# Polynomials (Algebra) - I.S.I. 2019 : Problem #7

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## Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] 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 $a_2=0$.

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I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 7.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.7" open="off"]Polynominals (Algebra)

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8 out of 10

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Polynomials - Edward Barbeau [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

## Start with hints

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Do you really need a hint? Try it first!

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

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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.

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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.

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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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## Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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## Similar Problem

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