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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