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.

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

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

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