A function $\psi \in L_2(\mathbb{R})$ is orthogonal to all integer shifts of a function $\varphi \in L_2(\mathbb{R})$ if and if only $$\sum_{k\in \mathbb{z}} \hat{\varphi}(\xi+k)\overline{\hat{\psi}(\xi+k)}\equiv0$$
My approach: $\int_{\mathbb{R}}\psi(t)\varphi(t+k)dt=0$ $\forall k \in \mathbb{Z}$ given. Now
\begin{align} &\sum_{k\in\mathbb{Z}}\int_{\mathbb{R}}\psi(t)\exp(-2\pi i (\xi+k)t)dt \overline{\int_{\mathbb{R}}\varphi(s)\exp(-2\pi i (\xi+k)s )}ds\\ \implies &\sum_{k\in\mathbb{Z}}\int_{\mathbb{R}}\psi(t)\exp(-2\pi i (\xi+k)t)dt \int_{\mathbb{R}}\varphi(s)\exp(2\pi i (\xi+k)s )ds\\ \implies &\sum_{k\in\mathbb{Z}}\int_{\mathbb{R}}\psi(t)\varphi(s)dsdt \int_{\mathbb{R}}\exp(2\pi i (\xi+k)s )\exp(-2\pi i (\xi+k)t)dtds\\ \end{align} The last step I am not so sure, I think this can be done if s and t are independent. I am open to further suggestions or corrections and ideas regarding how to proceed.
I got a hint from the prof that we need to use:
$$\varphi(\xi)=\sum_{k\in \mathbb{Z}}|\hat{\psi}(\xi+k)|^2$$
$\{\psi(t-k)\}_{k\in \mathbb{Z}}$ is a orthonormal system if and only if$\varphi\equiv1$