<s>[INST] Combinatorial sum inequality Prove the following inequality:
$$ \forall k\in\left\{4n+5:n\in\mathbb{N}\right\},\qquad\sum_{m=0}^{\frac{k-1}{2}}{\left( -1  \right)  }^{m}\binom{k}{2m}2^{2m}\neq 1. $$
I'm particularly interested in a combinatorial or number-theoretical solution.
 [/INST] We just have to compute:
$$\sum_{m=0}^{2n+2}\binom{4n+5}{2m}(2i)^{2m}=\frac{1}{2}\left(\sum_{k=0}^{4n+5}\binom{4n+5}{k}(2i)^k+\sum_{k=0}^{4n+5}\binom{4n+5}{k}(-2i)^k\right) $$
that is:
$$\sum_{m=0}^{2n+2}\binom{4n+5}{2m}(2i)^{2m}=\frac{(1+2i)^{4n+5}+(1-2i)^{4n+5}}{2}.$$
If we set:
$$ A_n = \frac{(1+2i)^{n}+(1-2i)^{n}}{2},\qquad B_n=A_{4n+5} $$
we have $A_0=1,A_1=1,B_0=-7,B_1=-527$ and:
$$ A_{n+2} = 2 A_{n+1} - 5 A_n,\qquad B_{n+2} = -14 B_{n+1}-625 B_n\equiv B_{n+1}\pmod{5},$$
so, by induction, the only possible residue class of $B_n\pmod{5}$ is $\color{red}{3}$ and $B_n$ cannot be $1$.
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