 # Understand the problem

Find all pairs $(b,c)$ of positive integers, such that the sequence defined by $a_1=b$, $a_2=c$ and $a_{n+2}= \left| 3a_{n+1}-2a_n \right|$ for $n \geq 1$ has only finite number of composite terms.

Number theory
Medium
##### Suggested Book
Problem Solving Strategies by Arthur Engel

Do you really need a hint? Try it first!

There exists a general theory of second-order linear difference equations. Read about it here.
Can we have $a_n>a_{n+1}$ for all $n$? What happens if we do? What happens otherwise?

Note that if $a_n for some $n$ then the sequence becomes increasing thereafter. Use this fact to simplify the recurrence relation and solve it explicitly.

The sequence cannot be decreasing because it is a sequence of positive integers. Hence there exists (a smallest) $k$ such that $a_k\le a_{k+1}$. If $a_k=a_{k+1}$ then the sequence becomes constant from the $k$th term onwards (we shall treat this case later). Otherwise $3a_{k+1}-2a_k>a_k$ hence $a_{k+2}>a_{k+1}$. This implies that the sequence becomes increasing from the $k$th term onwards. Also, $a_{n+2}=3a_{n+1}-2a_n$ for $n\ge k$. This difference equation has the characteristic equation $\lambda^2-3\lambda +2=0$ (see the link in hint 1) which has the solutions $\lambda = 2,1$. Thus, $a_{n+k}=2^nA+B$ for $A,B$ satisfying $A+B=a_k, 2A+B=a_{k+1}$. Take any prime divisor $p$ of $A+B$. By Fermat’s little theorem, $2^{m(p-1)} \equiv 1 \; (\text{mod}\; p)$ for every positive integer $m$. Thus $2^{m(p-1)}A+B\equiv A+B\equiv 0\; (\text{mod}\; p)$. Hence the sequence contains infinitely many composites. This cannot be allowed, so the sequence cannot be strictly increasing at any point.

The above discussion shows that, for any permissible sequence, there exists a (smallest) $j$ and a prime $q$ such that $a_n=q$ for all $n\ge j$. For $n, the sequence is decreasing. Note that, either $q=a_{j+1}=3a_j-2a_{j-1}=3q-2a_{j-1}$ or $q =2a_{j-1}-3q$. Hence, either $a_{j-1}=q$ or $a_{j-1}=2q$. The first one can happen only if $j=1$ because otherwise the minimality of $j$ is violated. In that case, the sequence is constant and $b=c=q$. If $a_{j-1}=2q$ then either $j=2$ (in which case $b=2q, c=q$) or $q=|6q-2a_{j-2}|$ hence $a_{j-2}=3q\pm\frac{q}{2}$. The last equality forces $q$ to be 2. Thus $a_{j-2}=6\pm 1$. If $j>3$ then $4=a_{j-1}=|18\pm 3 - 2a_{j-3}|$ which is absurd as $2a_{j-3}$ cannot be an odd number. Hence $j=3$ in this case and $c=4,b=5,7$.

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.

## Rectangular Piece of Paper | AMC 10A, 2014| Problem No 22

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2014. Problem-23. You may use sequential hints to solve the problem.

## Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability from AMC 10A, 2010. Problem-23. You may use sequential hints to solve the problem.

## Points on a circle | AMC 10A, 2010| Problem No 22

Try this beautiful Problem on Number theory based on Triangle and Circle from AMC 10A, 2010. Problem-22. You may use sequential hints to solve the problem.

## Circle and Equilateral Triangle | AMC 10A, 2017| Problem No 22

Try this beautiful Problem on Triangle and Circle from AMC 10A, 2017. Problem-22. You may use sequential hints to solve the problem.