# 1973 USAMO Problems/Problem 2

## 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.

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