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Consider the Somos 3 sequence $a_{n+3}a_n = a_{n+1}a_{n+2}$ with $a_1 = \alpha$, $a_2 = \beta$ and $a_3 = \gamma$ where $\alpha$, $\beta$ and $\gamma$ are all integers.

(a) Try different values of the initial data to see whether it generates integer sequences.

(b) From your observation, formulate the conditions on $\alpha$, $\beta$ and $\gamma$ which guarantee that the sequences generated are integers.

(c) Prove your statement formulated in (b).

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Append your work 'til now to the end of your question. – rschwieb Nov 7 '12 at 18:42

It's not too difficult to write out a few terms of your sequence. Let us fix $a_1,\ a_2,\ a_3$. Then your sequence begins $$a_1,\ a_2,\ a_3,\ a_4=\frac{a_3a_2}{a_1},\ a_5=\frac{a_3^2}{a_1},\ a_6=\frac{a_3^2a_2}{a_1^2},\ a_7=\frac{a_3^3}{a_1^2},\ a_8=\frac{a_3^3a_2}{a_1^3},\ a_9=\frac{a_3^4}{a_1^3}$$ The pattern should be rather clear. I will leave it to you to verify that the sequence continues in the expected manner. A necessary and sufficient condition is then $$\forall n\in\mathbb{N}:\ a_1^n\mid a_3^{n+1}\ \land\ a_1^n\mid a_3^na_2$$ I claim that $\forall n\in\mathbb{N}:\ a_1^n\mid a_3^{n+1}$ necessarily implies $a_1\mid a_3$. To see this, suppose otherwise.

Clearly the prime factors of $a_1$ is a subset of $a_3$'s. Therefore if $a_1\nmid a_3$ it must be because there is a prime factor occurring with larger exponent in $a_1$ than in $a_3$. Suppose that $p$ is such a prime with $p^k\|a_1$ and $p^\ell\|a_3$ such that $k > \ell$. But this is absurd, since $$a_1^n \mid a_3^{n+1} \implies p^{nk} \mid p^{(n+1)\ell} \implies nk < (n+1)\ell \implies \frac{k}{\ell} < 1 + \frac{1}{n}\ \ \forall n\in\mathbb{N}^+$$ The rational $\frac{k}{\ell}$ is bounded away from $1$ while the latter sequence converges towards $1$. This is clearly impossible.

Therefore it is necessary and sufficient that $a_1\mid a_3$.

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Let $b_n=\frac{a_{n+2}}{a_{n}}$, $$b_{n+1}=b_n=\cdots=b_1=\frac{a_{3}}{a_{1}}=\frac {\gamma}{\alpha}$$

$$\implies \frac{a_{n+2}}{a_{n}}=\frac {\gamma}{\alpha}$$

So, $a_{n+2}=\frac {\gamma}{\alpha}a_n=\cdots=\left(\frac {\gamma}{\alpha}\right)^{\frac{n+2-r}2} a_r$

If $n$ is even,$=2m$(say) $a_{2m+2}=\left(\frac {\gamma}{\alpha}\right)^{\frac{2m+2-r}2} a_r=\left(\frac {\gamma}{\alpha}\right)^ma_2=\left(\frac {\gamma}{\alpha}\right)^m\beta$

So, $(\alpha)^m\mid (\gamma)^m\beta$ to make $a_{2m+2}$ integer.

If $n$ is odd,$=2m+1$(say) $a_{2m+3}=\left(\frac {\gamma}{\alpha}\right)^{\frac{2m+3-r}2} a_r=\left(\frac {\gamma}{\alpha}\right)^ma_3=\left(\frac {\gamma}{\alpha}\right)^m\gamma$

So, $(\alpha)^m\mid (\gamma)^{m+1}$ to make $a_{2m+3}$ integer.

Combining $\alpha^m\mid (\gamma^{m+1},\gamma^m\beta)\iff\alpha^m\mid (\gamma^m(\beta,\gamma))$ to keep $a_r$ integer for integer $r\ge 4$

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Consider any prime $p$ and denote by $x_n$ the exponent with which $p$ is present in $a_n$. The recursion for the $a_n$ then implies $$x_{n+3} - x_{n+2}-x_{n+1}+x_n=0 \qquad(n\geq1)\ .\qquad(*)$$ The characteristic polynomial of this difference equation is $$\chi(\lambda)=\lambda^3-\lambda^2-\lambda+1=(\lambda-1)^2(\lambda+1)\ .$$ Therefore the general solution of $(*)$ is $$x_n=A \ n+B+ C(-1)^n\qquad (n\geq1)\ ,$$ whereby the coefficients $A$, $B$, $C$ have to be determined from the initial conditions.

If we want the $a_n$ to be integers for all $n\geq1$ then the exponents $x_1$, $x_2$, $x_3$ with which $p$ appears in the initial values $\alpha$, $\beta$, $\gamma$ have to be such that $x_n\geq0$ is guaranteed for all $n\geq 1$, and this for every prime $p$ individually.

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