$\displaystyle(1-\sqrt{2}x+x^2)^n=\sum_{k=0}^{2n}a_k x^k$
可證$a_{2n-k}=a_k$
$\displaystyle(1+\sqrt{2}x+x^2)^n=\sum_{k=0}^{2n}(-1)^k a_k x^k$
$\displaystyle(x^4+1)^n=\sum_{k=0}^n C_n^k x^{4k}=\sum_{k=0}^{4n}x^k(\sum_{m=0}^k (-1)^m a_{k-m}a_m)$
$4k=2n\Rightarrow k=\dfrac{n}{2}$
$\displaystyle\sum_{m=0}^{2n}(-1)^m a_m^2=\sum_{m=0}^{2n}(-1)^m a_{2n-m}a_m=
\begin{cases}
C_n^{n/2}~,2n\equiv 0\pmod{4}\\
0~,2n\equiv 2\pmod{4}
\end{cases}$ |