$\frac{a_{n+3}}{a_{n+1}}=\frac{a_{n+2}}{a_{n}}$
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 $r\ge 4$