The roots of a certain recursively-defined family of polynomials are all real Let $P_0=1 \,\text{and}\,P_1=x+1$ and we have $$P_{n+2}=P_{n+1}+xP_n\,\,n=0,1,2,...$$

Show that for all $n\in \mathbb{N}$, $P_n(x)$ has no complex root?

 A: For $x = u+iv$ define the pair of sequences $u_n,v_n$ such that $P_n(x) = u_n+iv_n$ are the real and imaginary parts of $P_n(x)$.
Then, $u_{n+1}+iv_{n+1} = u_n+iv_n + (u+iv)(u_{n-1}+iv_{n-1})$
$\implies u_{n+1} = u_n + uu_{n-1} - vv_{n-1}$ and $v_{n+1} = v_n + uv_{n-1} + vu_{n-1}$
Assume $a+ib$ (where, $a,b \in \mathbb{R}$ and $b \neq 0$) is a root of $P_n(x)$ i.e., $P_n(a+ib) = 0$ and say $m$ is the smallest index $n$ such that $P_m(a+ib) = 0$.
Since, the coefficients of $P_n$ are all real numbers, $P_m(a-ib) = 0$.
Thus, $u_m = 0 = u_{m-1}+au_{m-2}-bv_{m-2} = u_{m-1}+au_{m-2}+bv_{m-2}$
and $v_{m} = 0 = v_{m-1} + av_{m-2} + bu_{m-2} = v_{m-1} + av_{m-2} - bu_{m-2}$
That is $-bv_{m-2} = bv_{m-2}$ and $-bu_{m-2} = bu_{m-2}$ and since $b \neq 0$ we must also have 
$u_{m-2} = 0 = v_{m-2}$,i.e., $a+ib$ is a root of $P_{m-2}$ which contradicts the minimality of $m$.
Hence, $b = 0$, i.e., all the roots of $P_n$ are real.
Note: The roots of $P_n(x)$ are infact: $-\dfrac{1}{4}\sec^2 \frac{k\pi}{n+2}$ for $k = 1,2,\cdots \left[\frac{n+1}{2}\right]$.
A: Interlacing is a good hint, but let we show a brute-force solution. By setting:
$$ f(t) = \sum_{n\geq 0}P_n(x)\frac{t^n}{n!} \tag{1}$$
we have that the recursion translates into the ODE:
$$ f''(t) = f'(t) + x\, f(t) \tag{2}$$
whose solutions are given by:
$$ f(t) = A \exp\left(t\frac{1+\sqrt{1+4x}}{2}\right) + B\exp\left(t\frac{1-\sqrt{1+4x}}{2}\right)\tag{3}.$$
This gives:
$$ P_n(x) = A\left(\frac{1+\sqrt{1+4x}}{2}\right)^n + B\left(\frac{1-\sqrt{1+4x}}{2}\right)^n\tag{4}$$
and by plugging in the initial conditions we have:
$$ A=\frac{1+2x+\sqrt{1+4x}}{2\sqrt{1+4x}},\quad B=\frac{-1-2x+\sqrt{1+4x}}{2\sqrt{1+4x}}\tag{5}$$
so $P_n(x)=0$ is equivalent to:
$$\left(\frac{1+\sqrt{1+4x}}{1-\sqrt{1+4x}}\right)^n = \frac{1+2x-\sqrt{1+4x}}{1+2x+\sqrt{1+4x}},\tag{6}$$
or, by setting $x=\frac{y^2-1}{4}$, to:
$$\left(\frac{y+1}{1-y}\right)^n = \frac{y^2-y+1}{y^2+y+1} \tag{7} $$
or, by setting $y=\frac{z-1}{z+1}$, to:
$$ z^n = \frac{3+z^2}{1+3z^2}\tag{8} $$
or to:
$$ 3z^{n+2}+z^n-z^2-3 = 0.\tag{9} $$
