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Lemma If $x_0$ is a root of $P_n(x)$, then $0<x_0<1$. Lemma prove If $x_0\geqslant 1$, then $P_n(x_0)>0$ which is not true. If $x_0\leqslant 0$, let $y_0=1-x_0\geqslant 1$, then $P(x_0)=(-1)^nP_n(y_0)\neq 0$ which is not true. Prove For $\forall x_0\in (0,1)$, define $a=\sqrt{2x_0}$ and $b=\sqrt {1-x_0}$. Obviously $a,b>0$. $P_n(x_0)=\sum _{k=0}^n2^k\binom{2n}{2k}x_0^k(x_0-1)^{n-k}=\sum _{k=0}^n2^k\binom{2n}{2k}a^{2k}(ib)^{2n-2k}$ As $(a+ib)^{2n}=\sum _{k=0}^{2n}\binom{2n}{k}a^k(ib)^{2n-k}$, and $(a-ib)^{2n}=\sum _{k=0}^{2n}\binom{2n}{k}(-1)^ka^k(ib)^{2n-k}$. $P_n(x_0)=\frac 12\left[(a+ib)^{2n}+(a-ib)^{2n}\right]$. Notice that $a^2+b^2=x_0+1$, $\left(\dfrac{a}{\sqrt {x_0^2+1}}\right)^2+\left(\dfrac{b}{\sqrt {x_0^2+1}}\right)^2=1$. We can let $\dfrac{a}{\sqrt {x_0^2+1}}=\cos\varphi$, $\dfrac{b}{\sqrt {x_0^2+1}}=\sin\varphi(0<\varphi <\frac{\pi}{2})$. Therefore $P_n(x_0)=\frac 12\left[(\cos\varphi+i\sin\varphi )^{2n}+(\cos\varphi-i\sin\varphi)^{2n}\right]=\cos 2n\varphi$. $P_n(x_0)=0\Leftrightarrow\cos 2n\varphi =0\Leftrightarrow\varphi =\frac{\pi}{4n}+\frac{k\pi}{2n}(0\leqslant k\leqslant n-1)$. So there are $n$ different roots of $P_n(x)$.$\Box$