<s>[INST] if $x_{m}$be prime number, show that $2m+1$ must prime number! give the integer $a>1$, and define
$$x_{0}=1,x_{1}=4a+1,x_{n+1}=(4a+2)x_{n}-x_{n-1},n\ge 1$$
if $x_{m}$be prime number, show that $2m+1$ must prime number!
this problem seem interesting,Now  I post  follow my try : 
since Characteristic equation
$$r^2-2(2a+1)r=1\Longrightarrow r_{1,2}=2a+1+\sqrt{4a^2+4a},2a+1-\sqrt{4a^2+4a}$$
so we have
$$x_{n}=A(r_{1})^n+B(r_{2})^n$$
where
$$\begin{cases}
A+B=1\\
(2a+1)(A+B)+\sqrt{4a^2+4a}(A-B)=4a+1
\end{cases}$$
so we have 
$$A=\dfrac{\sqrt{a+1}+\sqrt{a}}{2\sqrt{a+1}},B=\dfrac{\sqrt{a+1}-\sqrt{a}}{2\sqrt{a+1}}$$
Note $$r_{1}=(\sqrt{a}+\sqrt{a+1})^2,r_{2}=(\sqrt{a+1}-\sqrt{a})^2$$
Let $$\alpha=\sqrt{a}+\sqrt{a+1},\beta=\sqrt{a+1}-\sqrt{a},\alpha\beta=1$$
so we have
$$x_{n}=\dfrac{\alpha^{2n+1}+\beta^{2n+1}}{\alpha+\beta}$$
 [/INST] As far as I can tell, everything you've shown is correct. As such, as you stated near the end of your question text, with
$$\alpha = \sqrt{a + 1} + \sqrt{a}, \; \; \beta = \sqrt{a + 1} - \sqrt{a}, \; \; \alpha\beta = 1 \tag{1}\label{eq1A}$$
you get
$$x_{n} = \frac{\alpha^{2n + 1} + \beta^{2n + 1}}{\alpha + \beta} \tag{2}\label{eq2A}$$
First, note for any integer $k \ge 0$ that
$$\begin{equation}\begin{aligned}
\alpha^{2k} + \beta^{2k} & = (\sqrt{a + 1} + \sqrt{a})^{2k} + (\sqrt{a + 1} - \sqrt{a})^{2k} \\
& = \sum_{i=0}^{2k}\binom{2k}{i}(\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i} + \sum_{i=0}^{2k}\binom{2k}{i}(\sqrt{a + 1})^{2k-i}(-\sqrt{a})^{i} \\
& = \sum_{i=0}^{2k}\binom{2k}{i}\left((\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i} + (-1)^i(\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i}\right) \\
& = \sum_{i=0}^{2k}\binom{2k}{i}(\sqrt{a + 1})^{2k-i}(\sqrt{a})^{i}(1 + (-1)^i) \\
& = 2\sum_{i=0}^{k}\binom{2k}{2i}(\sqrt{a + 1})^{2k-2i}(\sqrt{a})^{2i} \\
& = 2\sum_{i=0}^{k}\binom{2k}{2i}(a + 1)^{k-i}(a)^{i}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
The second last line comes from all of the terms with odd $i$ canceling as $1 + (-1)^{i} = 1 - 1 = 0$ and the even terms having a factor of $1 + (-1)^{i} = 1 + 1 = 2$. Also, since $a$ is an integer, this shows $\alpha^{2k} + \beta^{2k}$ is an integer.
Next, consider that $x_n$ in \eqref{eq2A} is a prime, but $2n + 1$ is not a prime. Since $x_0 = 1$ is not a prime, this means $n \ge 1$, so $2n + 1 \gt 1$ and, thus, must be composite. This means there are integers $q \gt 1$ and $r \gt 1$ where
$$2n + 1 = qr \tag{4}\label{eq4A}$$
Also, since $2n + 1$ is odd, this means $q$ and $r$ are also odd, so $q \ge 3$ and $r \ge 3$.
Note for all odd positive integers $s$, we have
$$\begin{equation}\begin{aligned}
x^s + y^s & = (x + y)(x^{s-1} - x^{s-2}y + \ldots - x(y^{s-2}) + y^{s-1}) \\
& = (x + y)\left(\sum_{i=0}^{s-1}(-1)^{i}x^{s-1-i}y^{i}\right)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Using \eqref{eq4A} and \eqref{eq5A}, then \eqref{eq2A} becomes
$$\begin{equation}\begin{aligned}
x_n & = \frac{\alpha^{qr} + \beta^{qr}}{\alpha + \beta} \\
& = \frac{(\alpha^{q})^{r} + (\beta^{q})^{r}}{\alpha + \beta} \\
& = \frac{(\alpha^{q} + \beta^{q})\left(\sum_{i=0}^{r-1}(-1)^{i}(\alpha^{q})^{r-1-i}(\beta^{q})^{i}\right)}{\alpha + \beta} \\
& = \left(\frac{\alpha^{q} + \beta^{q}}{\alpha + \beta}\right)\left(\sum_{i=0}^{r-1}(-1)^{i}(\alpha^{q})^{r-1-i}(\beta^{q})^{i}\right)
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
With the first factor, letting $q = 2m + 1$, we can see from \eqref{eq2A} it's $x_m$ and, thus, an integer. With the second factor, note the first & last terms are
$$(\alpha^{q})^{r-1} + (\beta^{q})^{r-1} \tag{7}\label{eq7A}$$
Since $r$ is odd, then $r - 1$ is even so by \eqref{eq3A} the value in \eqref{eq7A} is an integer. Next, consider the second term & the second last term of that factor,
$$\begin{equation}\begin{aligned}
-(\alpha^{q})^{r-2}(\beta^{q}) - (\alpha^{q})(\beta^{q})^{r-2} & = -(\alpha\beta)^{q}((\alpha^{q})^{r-3} + (\beta^{q})^{r-3}) \\
& = -((\alpha^{q})^{r-3} + (\beta^{q})^{r-3})
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
where $\alpha\beta = 1$ from \eqref{eq1A} was used. Once again, $r - 3$ is even so the value in \eqref{eq8A} is an integer.
You can repeat this pairing of terms from the start and end of the second factor in \eqref{eq6A} until you get to the middle term of
$$\begin{equation}\begin{aligned}
(-1)^{\frac{r-1}{2}}(\alpha^{q})^{\frac{r-1}{2}}(\beta^{q})^{\frac{r-1}{2}} & = (-1)^{\frac{r-1}{2}}\left(\alpha\beta\right)^{\frac{q(r-1)}{2}} \\
& = (-1)^{\frac{q(r-1)}{2}}
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$
This is also, of course, an integer. Thus, the second factor of \eqref{eq6A} is a sum of integers so it, too, is also an integer. In addition, you can easily show both these factors in \eqref{eq6A} are $\gt 1$. As such, this shows that $x_n$ is a product of $2$ integer factors each $\gt 1$ and, thus, it cannot be prime. However, since it was stated it was a prime, this is a contradiction. This means the assumption of $2n + 1$ being composite must be false, thus proving $2n + 1$ is actually a prime if $x_n$ is a prime.
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