Let $\{x_{n}: n\in \mathbb{Z}_{+}\}$ be a sequence of complex numbers such that $$\sum\limits_{p\in \mathbb{Z}_{+}}\vert x_{p} \vert^2=1$$ and
$$\sum\limits_{p\in \mathbb{Z}_{+} }x_{p} \overline{x_{p+r}}=0~~ \text{for all}~~ r\geq 1.$$ Does it imply at most finitely many $x_{p}$ can be non zero. Can you please provide ideas how to solve this? Any help is appreciated. Thank you in advance!
There is a counterexample to your question.
Set $x_{1}=1$.
For $n \geq 2$ set $x_{n} = q^{n-2}$ where $q = \frac{1-\sqrt{5}}{2}$
Note that $\sum_{j=1}^{\infty}|x_{j}|^2 = 1+\sum_{k=0}^{\infty}q^{2k} = 1+\frac{1}{1-q^2}= \frac{\sqrt{5}+3}{2} = 1+\phi$
Thus $(\frac{1}{\sqrt{1+\phi}}x_{j})_{j=1}^{\infty}$ is a normalized sequence.
We claim that $(\frac{1}{\sqrt{1+\phi}}x_{j})_{j=1}^{\infty}$ is a $\textbf{counterexample}$. We can ignore the normalization factor $\frac{1}{\sqrt{1+\phi}}$ for the computation below.
For $r \geq 1,$ $\sum_{j=1}^{\infty}x_{j}\overline{x_{j+r}}$ is
$$(q^{r-1})+\sum_{j=2}^{\infty}q^{j-2}q^{r+j-2}$$
$$ = q^{r-1} + \sum_{k=0}^{\infty}q^{r+2k}$$
$$ = q^{r-1}+q^{r}\frac{1}{1-q^{2}}$$
$$ = q^{r-1}(\frac{1-q^2+q}{1-q^2})$$
Note $1-q^2+q = 0$ by wolfram alpha, thus the above quantity is $0$.
$(\frac{1}{\sqrt{1+\phi}}x_{j})_{j=1}^{\infty}$ is a $\textbf{counterexample}$ to your question.