A sequence of (mostly) prime numbers

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]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.

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Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]There exists a general theory of second-order linear difference equations. Read about it here.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Can we have $latex a_n>a_{n+1}$ for all $latex n$? What happens if we do? What happens otherwise?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Note that if $latex a_n<a_{n+1}$ for some $latex n$ then the sequence becomes increasing thereafter. Use this fact to simplify the recurrence relation and solve it explicitly.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

The sequence cannot be decreasing because it is a sequence of positive integers. Hence there exists (a smallest) $latex k$ such that $latex a_k\le a_{k+1}$. If $latex a_k=a_{k+1}$ then the sequence becomes constant from the $latex k$th term onwards (we shall treat this case later). Otherwise $latex 3a_{k+1}-2a_k>a_k$ hence $latex a_{k+2}>a_{k+1}$. This implies that the sequence becomes increasing from the $latex k$th term onwards. Also, $latex a_{n+2}=3a_{n+1}-2a_n$ for $latex n\ge k$. This difference equation has the characteristic equation $latex \lambda^2-3\lambda +2=0$ (see the link in hint 1) which has the solutions $latex \lambda = 2,1$. Thus, $latex a_{n+k}=2^nA+B$ for $latex A,B$ satisfying $latex A+B=a_k, 2A+B=a_{k+1}$. Take any prime divisor $latex p$ of $latex A+B$. By Fermat's little theorem, $latex 2^{m(p-1)} \equiv 1 \; (\text{mod}\; p)$ for every positive integer $latex m$. Thus $latex 2^{m(p-1)}A+B\equiv A+B\equiv 0\; (\text{mod}\; p)$. 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) $latex j$ and a prime $latex q$ such that $latex a_n=q$ for all $latex n\ge j$. For $latex n<j$, the sequence is decreasing. Note that, either $latex q=a_{j+1}=3a_j-2a_{j-1}=3q-2a_{j-1}$ or $latex q =2a_{j-1}-3q$. Hence, either $latex a_{j-1}=q$ or $latex a_{j-1}=2q$. The first one can happen only if $latex j=1$ because otherwise the minimality of $latex j$ is violated. In that case, the sequence is constant and $latex b=c=q$. If $latex a_{j-1}=2q$ then either $latex j=2$ (in which case $latex b=2q, c=q$) or $latex q=|6q-2a_{j-2}|$ hence $latex a_{j-2}=3q\pm\frac{q}{2}$. The last equality forces $latex q$ to be 2. Thus $latex a_{j-2}=6\pm 1$. If $latex j>3$ then $latex 4=a_{j-1}=|18\pm 3 - 2a_{j-3}|$ which is absurd as $latex 2a_{j-3}$ cannot be an odd number. Hence $latex j=3$ in this case and $latex c=4,b=5,7$.

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Watch the video (Coming Soon)

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Similar Problems

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