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Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.

We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n-2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0<t<\pi/2$ and $$h(t):=Q_n(\tan^2 t)=\frac{\cos nt}{\cos^n t}-n\tan^2 t=\frac{\cos nt-n\sin^2 t\cos^{n-2}t}{\cos^n t}.$$

Denote $a=n/2-1$ and note that the maximal value $(1-x)x^a$ over $x\in [0,1]$ is obtained for $x=a/(a+1)$ and equals $\frac{1}{(a+1)(1+1/a)^a}<\frac{1}{2(a+1)}=\frac1n$, applying this for $x=\cos^2 t$ we see that $0\leqslant n\sin^2 t\cos^{n-2}t<1$, thus the signs of $h(t)$ at the points $\pi k/n$, $k=0, 1,\ldots, \lfloor n/2\rfloor$ interchange and $h$ has at least $\lfloor n/2\rfloor$ distinct roots on $(0,\pi/2)$, as desired.

Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.

We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n+2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0<t<\pi/2$ and $$h(t):=Q_n(\tan^2 t)=\frac{\cos nt}{\cos^n t}+n\tan^2 t=\frac{\cos nt+n\sin^2 t\cos^{n-2}t}{\cos^n t}.$$

Denote $a=n/2-1$ and note that the maximal value $(1-x)x^a$ over $x\in [0,1]$ is obtained for $x=a/(a+1)$ and equals $\frac{1}{(a+1)(1+1/a)^a}<\frac{1}{2(a+1)}=\frac1n$, applying this for $x=\cos^2 t$ we see that $0\leqslant n\sin^2 t\cos^{n-2}t<1$, thus the signs of $h(t)$ at the points $\pi k/n$, $k=0, 1,\ldots, \lfloor n/2\rfloor$ interchange and $h$ has at least $\lfloor n/2\rfloor$ distinct roots on $(0,\pi/2)$, as desired.

Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.

We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n-2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0<t<\pi/2$ and $$h(t):=Q_n(\tan^2 t)=\frac{\cos nt}{\cos^n t}-n\tan^2 t=\frac{\cos nt-n\sin^2 t\cos^{n-2}t}{\cos^n t}.$$

Denote $a=n/2-1$ and note that the maximal value $(1-x)x^a$ over $x\in [0,1]$ is obtained for $x=a/(a+1)$ and equals $\frac{1}{(a+1)(1+1/a)^a}<\frac{1}{2(a+1)}=\frac1n$, applying this for $x=\cos^2 t$ we see that $0\leqslant n\sin^2 t\cos^{n-2}t<1$, this the signs of $h(t)$ at the points $\pi k/n$, $k=0, 1,\ldots, \lfloor n/2\rfloor$ interchange and $h$ has at least $\lfloor n/2\rfloor$ distinct roots on $(0,\pi/2)$, as desired.

Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.

We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n-2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0<t<\pi/2$ and $$h(t):=Q_n(\tan^2 t)=\frac{\cos nt}{\cos^n t}-n\tan^2 t=\frac{\cos nt-n\sin^2 t\cos^{n-2}t}{\cos^n t}.$$

Denote $a=n/2-1$ and note that the maximal value $(1-x)x^a$ over $x\in [0,1]$ is obtained for $x=a/(a+1)$ and equals $\frac{1}{(a+1)(1+1/a)^a}<\frac{1}{2(a+1)}=\frac1n$, applying this for $x=\cos^2 t$ we see that $0\leqslant n\sin^2 t\cos^{n-2}t<1$, thus the signs of $h(t)$ at the points $\pi k/n$, $k=0, 1,\ldots, \lfloor n/2\rfloor$ interchange and $h$ has at least $\lfloor n/2\rfloor$ distinct roots on $(0,\pi/2)$, as desired.

Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.

We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n-2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0<t<\pi/2$ and $$h(t):=Q_n(\tan^2 t)=\frac{\cos nt}{\cos^n t}-n\tan^2 t=\frac{\cos nt-n\sin^2 t\cos^{n-2}t}{\cos^n t}.$$

Denote $a=n/2-1$ and note that the maximal value $(1-x)x^a$ over $x\in [0,1]$ is obtained for $x=a/(a+1)$ and equals $\frac{1}{(a+1)(1+1/a)^a}<\frac{1}{2(a+1)}=\frac1n$, applying this for $x=\cos^2 t$ we see that $\cos nt-n\sin^2 t\cos^{n-2}t>0$ when $\cos nt=1$ and $\cos nt-n\sin^2 t\cos^{n-2}t<0$ when $\cos nt=0$. This already gives enough many sign changes of $\cos nt-n\sin^2 t\cos^{n-2}t$.

Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.

We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n-2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0<t<\pi/2$ and $$h(t):=Q_n(\tan^2 t)=\frac{\cos nt}{\cos^n t}-n\tan^2 t=\frac{\cos nt-n\sin^2 t\cos^{n-2}t}{\cos^n t}.$$

Denote $a=n/2-1$ and note that the maximal value $(1-x)x^a$ over $x\in [0,1]$ is obtained for $x=a/(a+1)$ and equals $\frac{1}{(a+1)(1+1/a)^a}<\frac{1}{2(a+1)}=\frac1n$, applying this for $x=\cos^2 t$ we see that $0\leqslant n\sin^2 t\cos^{n-2}t<1$, this the signs of $h(t)$ at the points $\pi k/n$, $k=0, 1,\ldots, \lfloor n/2\rfloor$ interchange and $h$ has at least $\lfloor n/2\rfloor$ distinct roots on $(0,\pi/2)$, as desired.