Summing even and odd terms of binomial coefficients, with increasing powers of 2 and alternating signs For a complex number $z$, let $(1+z)^{15}=a_{0}+a_{1} z+\cdots+a_{15} z^{15}$. The value of
$$
\left(a_{0}-4 a_{2}+16 a_{4}+\cdots-2^{14} a_{14}\right)^{2}+\left(2 a_{1}-8 a_{3}+32 a_{5} \cdots-2^{15} a_{15}\right)^{2}
$$
While I was entirely stumped by this question, I found some similar questions on here, specifically A sum involving binomial coefficients, powers and alternating signs. This did not equip me to be able to solve the above question, though, since this problem drops odd terms and even terms, alternates signs, AND has a factor of increasing powers of $2$, not to mention squares the result and then sums them.
I don't see how the sum can even be expressed as a binomial to somehow use calculus, nor do I see any algebraic manipulations; I've only managed to embarrassingly restate the problem statement as
$$ \left(\sum_{k=0}^{7} 2^{2 k}({ }^{15} C_{2 k})(-1)^k\right)^{2}+\left(\sum_{k=0}^{7} 2^{2 k+1}\left({ }^{15} C_{2 k+1}\right)(-1)^k\right)^{2} $$

A slight variation. We consider the polynomial
\begin{align*}
\color{blue}{F(z)}&\color{blue}{=\left(\sum_{k=0}^7\binom{15}{2k}(-1)^k(2z)^{2k}\right)^2+\left(\sum_{k=0}^7\binom{15}{2k+1}(-1)^k(2z)^{2k+1}\right)^2}\tag{1}
\end{align*}
and with respect to OPs expression we need to calculate $F(1)$.

We obtain from (1)
\begin{align*}
\color{blue}{F(z)}&=\left(\sum_{{k=0}\atop{k\mathrm{\ even}}}^{15}\binom{15}{k}i^k(2z)^{k}\right)^2
+\left(\frac{1}{i}\sum_{{k=0}\atop{k\mathrm{\ odd}}}^{15}\binom{15}{k}i^k(2z)^{k}\right)^2\\
&=\left(\frac{(1+2iz)^{15}+(1-2iz)^{15}}{2}\right)^2+\left(\frac{1}{i}\,\frac{(1+2iz)^{15}-(1-2iz)^{15}}{2}\right)^2\\
&=\frac{1}{4}\left((1+2iz)^{15}+(1-2iz)^{15}\right)^2-\frac{1}{4}\left((1+2iz)^{15}-(1-2iz)^{15}\right)^2\\
&=\left(1+2iz\right)^{15}\left(1-2iz\right)^{15}\\
&\,\,\color{blue}{=\left(1+4z^2\right)^{15}}\\\\
\color{blue}{F(1)}&=\color{blue}{5^{15}}
\end{align*}
in accordance with another given answer.

