Consider a real number $b$ and define the sequence of polynomials $P_0=1$, $P_1=x$, and for $n \ge 2$, $P_n=xP_{n-1}-b^2P_{n-2}$.

I conjecture that for $n \ge 1$, $P_n$ has $n$ distinct real roots.

One way to do this would be, for a fixed $x_0$, to note that $P_n(x_0)$ satisfies a linear recurrence of order $2$, and is of the form $Ar_1^n+Br_2^n$. Hence we can set this equal to zero and by introducing a root of unity $\omega$, we find a quadratic equation which $x_0$ must satisfy. This does work but it's a bit messy.

Is there maybe a nicer way?

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    $\begingroup$ You can check this link de.wikipedia.org/wiki/Tridiagonal-Toeplitz-Matrix and note that your recurrence relation is the same by finding the appropriate transformation. $\endgroup$ – Gregory Nov 27 '17 at 17:29
  • $\begingroup$ Or just note that your recurrence relation is the same as the eigenvalues for a tridiagonal toeplitz matrix, which in your case is also symmetric and symmetric matrices have real eigenvalues hence real roots. $\endgroup$ – Gregory Nov 27 '17 at 17:29
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    $\begingroup$ That's a known fact. The polynomials form a Sturm chain, the recursion being an explicit Euclid's algorithm. Actually, they are just scaled Chebyshev polynomials. $\endgroup$ – Professor Vector Nov 27 '17 at 17:44

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