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Prove that there exist infinitely many positive integers $k$ such that the sequence $\{x_n\}$ satisfying

$$ x_1=1, x_2=k+2, x_{n+2}-(k+1)x_{n+1}+x_n=0(n \ge 0)$$ does not contain any prime number.

I found a similar question: The sequence ${x_n}$ satisfies $x_{n+2}=(k+1)x_{n+1}-x_n, x_0=1, x_1=k+2$ and every term of the sequence is either $1$ or composite. Prove that the set of such $k$ is infinite.

Solution: Let $\alpha=\frac{\sqrt{k+3}+\sqrt{k-1}}{2}$, $\beta=\frac{\sqrt{k+3}-\sqrt{k-1}}{2}$, then$$x_n=\frac{\alpha^{2n+1}-\beta^{2n+1}}{\alpha-\beta}.$$We consider$$x_{n}^2=\frac{(\alpha^{2n+1}+\beta^{2n+1})^2-4}{k-1},$$let $a_n=\alpha^{2n+1}+\beta^{2n+1}$, $a_0=\sqrt{k+3}$, $a_1=\frac{(k+3)^{\frac{3}{2}}+3\sqrt{k+3}(k-1)}{4}$. We let $k+3=u^2$, and $2|u$, then $a_n$ is a positive integer, we have$$(k-1)x_{n}^2=(a_{n}+2)(a_{n}-2).$$If $x_n=p$ is a prime, $p$ is lagre enough when $n$ is lagre enough, $gcd(a_n-2, a_n+2)=gcd(a_n-2, 4)\leq 4$, it follows that $p^2\mid a_n+2$, or $p^2\mid a_n-2$, a contradiction.

Is a different solution like this?

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  • $\begingroup$ What you call "a similar question" is actually identical, isn't it? $\endgroup$ – Gerry Myerson Nov 19 '19 at 0:54
  • $\begingroup$ @GerryMyerson I agree, I think he wants another solution. If so, edit your question please! $\endgroup$ – user725958 Nov 19 '19 at 1:01
  • $\begingroup$ Hint: Try using the Z-transform. $\endgroup$ – Dinno Koluh Nov 19 '19 at 1:11
  • $\begingroup$ I tried to workout the question for $k=1$ and I got that $x_n = 2n-1$ which contains prime numbers. $\endgroup$ – Dinno Koluh Nov 19 '19 at 1:14
  • $\begingroup$ @DinnoKoluh What did you see? $\endgroup$ – trombho Nov 19 '19 at 1:17
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I tried using the Z transform. From the question, we conclude that $x_0 = -1$. And taking the Z transform we get: $$ x_{n+2} - (k+1)x_{n+1}+x_n = 0 $$ $$ z^2X(z)-x_0z^2-x_1z-(k+1)(zX(z)-x_0z) + X(z) = 0 $$ $$ X(z) = \frac{(k+1)z+z-z^2}{z^2-(k+1)z+1} $$ if we let $a=k+1$ then after taking the inverse Z transform we get: $$ x_n = 2^{-n-1}\frac{ (-\sqrt{a^2-4}+a+2)(\sqrt{a^2-4}+a)^n - (\sqrt{a^2-4}+a+2)(a-\sqrt{a^2-4})^n }{\sqrt{a^2-4}}$$ This might be helpful for further solving.

For $a=2$ which is $k=1$ the sequence becomes: $$ x_n = 2n - 1 $$ and it contains infinitely many prime numbers since it is a sequence of odd numbers and we have a contradiction.

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  • $\begingroup$ But the problem was to show there exists $k$ such that there are no primes, not to show that for all $k$ there are no primes. $\endgroup$ – Gerry Myerson Nov 19 '19 at 2:04
  • $\begingroup$ I understand that, so I wrote that this might be helpful to solve the problem not that it's a solution. $\endgroup$ – Dinno Koluh Nov 19 '19 at 2:06

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