多项式的奇/偶次系数的交替和的平方和

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Hall and Knight - Higher Algebra page 489:

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If $a_1,a_2,...,a_n$ are the roots of the equation $x^n+p_1x^{n-1}+p_2x^{n-2}+...+p_{n-1}x+p_n=0$ ($p_i$ are real)
prove that$$ (1+{a_1}^2)(1+{a_2}^2)...(1+{a_n}^2)=(1-p_2+p_4-...)^2+(p_1-p_3+p_5-...)^2$$应该可以这样解:
\begin{align*}
f(i)f(-i)&=(i^n+p_1i^{n-1}+p_2i^{n-2}+\cdots)((-i)^n+p_1(-i)^{n-1}+p_2(-i)^{n-2}+\cdots)\\
&=(1-p_1i+p_2i^2+\cdots)(1+p_1i+p_2i^2+\cdots)\\\boxed{\small a^2-b^2=(a+b)(a-b)}
&=(1+p_2i^2+p_4i^4+\cdots)^2-(p_1i+p_3i^3+p_5i^5+\cdots)^2\\
&=(1-p_2+p_4-\cdots)^2+(p_1-p_3+p_5-\cdots)^2
\end{align*}
另一方面
\begin{align*}
f(x)&=(x-a_1)\dots(x-a_n)\\
\Rightarrow f(i)f(-i)&=(i-a_1)\cdots(i-a_n)(-i-a_1)\cdots(-i-a_n)\\
&=(a_1^2-i^2)\cdots(a_n^2-i^2)\\
&=(a_1^2+1)\cdots(a_n^2+1)\\
\end{align*}

kuing

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