Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1973 USAMO Problems/Problem 2: Difference between revisions

← Older edit
Juicefruit (talk | contribs)
editor
34 edits
Ddk001 (talk | contribs)
editor
979 edits
No edit summary
 
Line 31: Line 31:




{{alternate solutions}}


==Solution 2==
==Solution 2==
Line 70: Line 67:


Therefore, the left hand side is always increasing and will never equal 5 again, so equality between <math>X_n</math> and <math>Y_n</math> only holds when <math>n = 0</math>, so they only share 1 term.
Therefore, the left hand side is always increasing and will never equal 5 again, so equality between <math>X_n</math> and <math>Y_n</math> only holds when <math>n = 0</math>, so they only share 1 term.
==Solution 3 (simple)==
First, <math>X_n</math> and <math>Y_n</math> are both positive for all nonnegative integers <math>n</math>. This is since <math>X_1</math>, <math>X_2</math>, <math>Y_1</math>, and <math>Y_2</math> are all positive, and if <math>X_n</math> and <math>X_{n-1}</math> is positive then <math>X_{n+1}</math> as a sum of two positive numbers is positive, and similarly for <math>Y_n</math>. Simple.
Next, consider <math>Z_n = Y_n - X_n</math>. We have <math>Z_{n+1} = Y_n + Z_n + 3 Z_{n-1} + X_{n-1}</math>. Another easy inductive argument like above shows that <math>Z_n >0</math> if <math>n>0</math>. But if <math>Z_n >0</math> then <math>Z_n \ne 0</math>, so <math>X_n \ne Y_n</math> as desired.
~ [[Ddk001]]
{{alternate solutions}}


==See Also==
==See Also==

Latest revision as of 21:09, 28 October 2025

Problem

Let $\{X_n\}$ and $\{Y_n\}$ denote two sequences of integers defined as follows:

$X_0=1$, $X_1=1$, $X_{n+1}=X_n+2X_{n-1}$ $(n=1,2,3,\dots),$
$Y_0=1$, $Y_1=7$, $Y_{n+1}=2Y_n+3Y_{n-1}$ $(n=1,2,3,\dots)$.

Thus, the first few terms of the sequences are:

$X:1, 1, 3, 5, 11, 21, \dots$,
$Y:1, 7, 17, 55, 161, 487, \dots$.

Prove that, except for the "1", there is no term which occurs in both sequences.

Solution

We can look at each sequence $\bmod{8}$:

$X$: $1$, $1$, $3$, $5$, $3$, $5$, $\dots$,
$Y$: $1$, $7$, $1$, $7$, $1$, $7$, $\dots$.
  • Proof that $X$ repeats $\bmod{8}$:

The third and fourth terms are $3$ and $5$ $\bmod{8}$. Plugging into the formula, we see that the next term is $11\equiv 3\bmod{8}$, and plugging $5$ and $3$, we get that the next term is $13\equiv 5\bmod{8}$. Thus the sequence $X$ repeats, and the pattern is $3,5,3,5,\dots$.

  • Proof that $Y$ repeats $\bmod{8}$:

The first and second terms are $1$ and $7$ $\bmod{8}$. Plugging into the formula, we see that the next term is $17\equiv 1\bmod{8}$, and plugging $7$ and $1$, we get that the next term is $23\equiv 7\bmod{8}$. Thus the sequence $Y$ repeats, and the pattern is $1,7,1,7,\dots$.


Combining both results, we see that $X_i$ and $Y_j$ are not congruent $\bmod{8}$ when $i\geq 3$ and $j\geq 2$. Thus after the "1", the terms of each sequence are not equal.


Solution 2

We can solve this problem by finding a particular solution for each linear recurrence.

$X_{n+1} - X_n - 2X_{n-1} = 0$ with characteristic polynomial

$X^2 - X - 2 = (X - 2)(X + 1)$, so $X = -1, 2$

$X_n = A(2)^n + B(-1)^n$

After plugging in to find the particular solution:

$X_0 = 1 = A + B$

$X_1 = 1 = 2A - B$

$A = \frac{1}{3}$ and $B = \frac{2}{3}$, so

$X_n = \frac{1}{3}(2)^n + \frac{2}{3}(-1)^n$

Doing the same for $Y_n$, we get

$Y_n = 2(3)^n - (-1)^n$

We know they're equal at $n = 0$, so let's set them equal and compare.

$2(3)^n - (-1)^n = \frac{1}{3}((2)^n + 2(-1)^n)$

$6(3)^n - 3(-1)^n = (2)^n + 2(-1)^n$

$6(3)^n - 2^n = 5(-1)^n$

By induction, we know in general that $(\alpha + 1)^n > \alpha^n$ for all $\alpha, n > 0$, so $6(3)^n > 2^n$ for all $n > 1$.

Therefore, the left hand side is always increasing and will never equal 5 again, so equality between $X_n$ and $Y_n$ only holds when $n = 0$, so they only share 1 term.


Solution 3 (simple)

First, $X_n$ and $Y_n$ are both positive for all nonnegative integers $n$. This is since $X_1$, $X_2$, $Y_1$, and $Y_2$ are all positive, and if $X_n$ and $X_{n-1}$ is positive then $X_{n+1}$ as a sum of two positive numbers is positive, and similarly for $Y_n$. Simple.

Next, consider $Z_n = Y_n - X_n$. We have $Z_{n+1} = Y_n + Z_n + 3 Z_{n-1} + X_{n-1}$. Another easy inductive argument like above shows that $Z_n >0$ if $n>0$. But if $Z_n >0$ then $Z_n \ne 0$, so $X_n \ne Y_n$ as desired.


~ Ddk001


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1973 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1973_USAMO_Problems/Problem_2&oldid=264366"