Here is the problem I can't solve.

Let $\xi_n$ $(n=1,2,3,\dots)$ be a sequence of i.i.d. random variables on $\mathbb{R}$ with density $p(x)>0$, let $\eta_n=\sum_{i=1}^{n}\xi_i^2$. Define $$\zeta_t = \eta_{[t]}(t-[t]) \times \eta_{[t]+1}(-t+[t]+1),$$ where $[t]$ denotes the integer part of $t$. Is $\zeta_t$ a Markov process?

I tried to prove that given process is Markov by definition: $$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t}) =\mathbb{P}(\zeta_{t+s}\in A| \mathcal{F_{=t}}),$$ where $A \in \mathcal{F}_{\ge t}$, $\mathcal{F}_{\le t} = \sigma(\zeta_s: s\le t)$, $\mathcal{F}_{\ge t} = \sigma(\zeta_s: s\ge t)$, $\mathcal{F}_{= t} = \sigma(\zeta_t)$.

Here is my first attempt so far

$$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t})=\mathbb{P}(\eta_{[t+s]}(t+s-[t+s]) \times \eta_{[t+s]+1}(-t-s+[t+s]+1)\in A|\mathcal{F}_{\le t})=$$
$$=\mathbb{P}\left(\left(\sum_{k=1}^{[t+s]}\xi_k^2(t+s-[t+s])\right)\times\left(\sum_{i=1}^{[t+s]+1}\xi_i^2([t+s]+1-t-s)\right)\in A|\mathcal{F}_{\le t}\right)=$$

$$=\sum_{k=1}^{[t+s]}\sum_{i=1}^{[t+s]+1}\mathbb{E}(\mathbb{1}_A\xi_k^2(t+s-[t+s])\xi_i^2([t+s]+1-t-s)|\mathcal{F}_{\le t}),$$ using that $\mathbb{P}(A|\mathcal{F})=\mathbb{E}(\mathbb{1}_A|\mathcal{F})$.

My second attempt was to make an example of suitable random variables and probability space, such that $\zeta_t$ is not Markov. I began to think it is not Markov as $\zeta_t$ depends on $\eta_{[t]+1}$, so it depends on the future in some sense because $[t]+1>t$.

I understand that it is kind of "not Mathoverflow question", but I did not receive answer on MathStackexchange and I faced this problem during my term paper, so may be this is a reason to post it here.