<s>[INST] 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?

 [/INST] 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 $k = 1,2,\cdots \left[\frac{n+1}{2}\right]$.
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