# Prime-free sequence

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?

• What you call "a similar question" is actually identical, isn't it? – Gerry Myerson Nov 19 '19 at 0:54
• @GerryMyerson I agree, I think he wants another solution. If so, edit your question please! – user725958 Nov 19 '19 at 1:01
• Hint: Try using the Z-transform. – Dinno Koluh Nov 19 '19 at 1:11
• I tried to workout the question for $k=1$ and I got that $x_n = 2n-1$ which contains prime numbers. – Dinno Koluh Nov 19 '19 at 1:14
• @DinnoKoluh What did you see? – trombho Nov 19 '19 at 1:17

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.
• 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. – Gerry Myerson Nov 19 '19 at 2:04