2020 CHKMO #1

 Given that ${a_n}$ and ${b_n}$ are two sequences of integers defined by

$$a_1=1, a_2=10, a_{n+1}=2a_n+3a_{n-1}, \dots \text{for }n=2,3,4,\ldots,$$

$$b_1=1, b_2=8, b_{n+1}=3b_n+4b_{n-1}, \dots \text{for }n=2,3,4,\ldots.$$

Prove that, besides the number $1$, no two numbers in the sequences are identical.



Using the standard recurrence relations to explicit equation, we compute the explicit formulas for each sequence:


$$a_n = \frac{11 \cdot 3^n + 21(-1)^n}{12}$$

$$b_n = \frac{9 \cdot 4^n + 16 (-1)^n }{20}$$

Thus it is enough to show that the equation

\[ 55 \cdot 3^n + 105(-1)^n = 27 \cdot 4^m + 48(-1)^m  \]has no solutions over the integers. Assume otherwise for the sake of contradiction. One can easily check that there is no $b_i=10=a_2$. Henceforth we assume $n > 2$. Then taking mod 27, we obtain

\[  105 \equiv \pm 48 \pmod{27} \]which is false.

Comments

Popular posts from this blog

1995 IMO #2

Inequality problem I made