2018 Putnam B Problems/Problem 4: Difference between revisions
Pinotation (talk | contribs) |
Pinotation (talk | contribs) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 27: | Line 27: | ||
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. | 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. | ||
So, 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. | 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 | ~Pinotation | ||
==See Also== | |||
[[2018 Putnam B Problems|2018 Putnam B Entire Test]] | |||
[[2018 Putnam B Problems/Problem 3|2018 Putnam B Problem 3]] | |||
[[2018 Putnam B Problems/Problem 5|2018 Putnam B Problem 5]] | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:23, 18 August 2025
Problem
Given a real number 
, we define a sequence by 
, 
, and 
for 
. Prove that if 
for some 
, then the sequence is periodic.
Solution
Suppose \( x_m = 0 \) for some \( m \geq 0 \). Then using the recurrence
![\[x_{n+1} = 2x_n x_{n-1} - x_{n-2},\]](http://latex.artofproblemsolving.com/d/7/6/d7695980452ce78cf896ab02f2d80b0d937d1e58.png)
we can compute the next few terms explicitly in terms of \( x_{m-2} \) and \( x_{m-1} \):
![\[x_{m+1} = 2x_m x_{m-1} - x_{m-2} = -x_{m-2},\]](http://latex.artofproblemsolving.com/c/5/9/c5977e68fde77845e60872074c5807c8bba0509f.png)
![\[x_{m+2} = 2x_{m+1} x_m - x_{m-1} = -x_{m-1},\]](http://latex.artofproblemsolving.com/0/2/8/028cf0916a4f4b19c738bec8bf8fa0b6f8f43f6a.png)
![\[x_{m+3} = 2x_{m+2} x_{m+1} - x_m = 2(-x_{m-1})(-x_{m-2}) - 0 = 2x_{m-1} x_{m-2}.\]](http://latex.artofproblemsolving.com/b/a/2/ba295cf96f96ccb05105e15f5ba5e5e5810203f4.png)
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
![\[(x_{m-1}, x_m), \quad (x_m, x_{m+1}), \quad (x_{m+1}, x_{m+2}), \ldots\]](http://latex.artofproblemsolving.com/2/d/b/2db1407fccac1f7308a5ae9609d79f9b7c379132.png)
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.
So, 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
See Also
These problems are copyrighted © by the Mathematical Association of America. 
