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Difference between revisions of "2018 Putnam B Problems/Problem 4"

(Created page with "==Problem== Given a real number <math>a</math>, we define a sequence by <math>x_0 = 1</math>, <math>x_1 = x_2 = a</math>, and <math>x_{n+1} = 2x_nx_{n-1} - x_{n-2}</math> for...")
 
(→‎Solution)
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==Solution==
 
==Solution==
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Suppose \( x_m = 0 \) for some \( m \geq 0 \). Then using the recurrence
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<cmath>
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x_{n+1} = 2x_n x_{n-1} - x_{n-2},
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</cmath>
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we can compute the next few terms explicitly in terms of \( x_{m-2} \) and \( x_{m-1} \):
 +
<cmath>
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x_{m+1} = 2x_m x_{m-1} - x_{m-2} = -x_{m-2},
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</cmath>
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<cmath>
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x_{m+2} = 2x_{m+1} x_m - x_{m-1} = -x_{m-1},
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</cmath>
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<cmath>
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x_{m+3} = 2x_{m+2} x_{m+1} - x_m = 2(-x_{m-1})(-x_{m-2}) - 0 = 2x_{m-1} x_{m-2}.
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</cmath>
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Continuing in this way, each term can be expressed as a polynomial in \( x_{m-2} \) and \( x_{m-1} \) with integer coefficients. In particular, the sequence starting from \( x_m = 0 \) depends only on the fixed pair \( (x_{m-2}, x_{m-1}) \).
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Since the pair \( (x_{m-2}, x_{m-1}) \) is fixed, the sequence of consecutive pairs
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<cmath>
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(x_{m-1}, x_m), \quad (x_m, x_{m+1}), \quad (x_{m+1}, x_{m+2}), \ldots
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</cmath>
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can take only finitely many distinct values determined algebraically from \( (x_{m-2}, x_{m-1}) \). Therefore, by the Pigeonhole Principle, some pair must eventually repeat. Once a pair repeats, the recurrence determines all subsequent terms uniquely, so the sequence becomes periodic.
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Therefore, if \( x_n = 0 \) for some \( n \), the sequence is periodic.
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Note: I saw this problem also had no solution so here it is. I am new to AoPS so if there is anything I am missing please edit this.
 +
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~Pinotation
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{{MAA Notice}}

Revision as of 20:51, 17 August 2025

Problem

Given a real number $a$, we define a sequence by $x_0 = 1$, $x_1 = x_2 = a$, and $x_{n+1} = 2x_nx_{n-1} - x_{n-2}$ for $n \ge 2$. Prove that if $x_n = 0$ for some $n$, then the sequence is periodic.

Solution

Suppose \( x_m = 0 \) for some \( m \geq 0 \). Then using the recurrence \[x_{n+1} = 2x_n x_{n-1} - x_{n-2},\] we can compute the next few terms explicitly in terms of \( x_{m-2} \) and \( x_{m-1} \): \[x_{m+1} = 2x_m x_{m-1} - x_{m-2} = -x_{m-2},\] \[x_{m+2} = 2x_{m+1} x_m - x_{m-1} = -x_{m-1},\] \[x_{m+3} = 2x_{m+2} x_{m+1} - x_m = 2(-x_{m-1})(-x_{m-2}) - 0 = 2x_{m-1} x_{m-2}.\] Continuing in this way, each term can be expressed as a polynomial in \( x_{m-2} \) and \( x_{m-1} \) with integer coefficients. In particular, the sequence starting from \( x_m = 0 \) depends only on the fixed pair \( (x_{m-2}, x_{m-1}) \).

Since the pair \( (x_{m-2}, x_{m-1}) \) is fixed, the sequence of consecutive pairs \[(x_{m-1}, x_m), \quad (x_m, x_{m+1}), \quad (x_{m+1}, x_{m+2}), \ldots\] can take only finitely many distinct values determined algebraically from \( (x_{m-2}, x_{m-1}) \). Therefore, by the Pigeonhole Principle, some pair must eventually repeat. Once a pair repeats, the recurrence determines all subsequent terms uniquely, so the sequence becomes periodic.

Therefore, if \( x_n = 0 \) for some \( n \), the sequence is periodic.

Note: I saw this problem also had no solution so here it is. I am new to AoPS so if there is anything I am missing please edit this.

~Pinotation

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