This is a problem from Durrett's probability with examples, exercise 8.2.1. It is not homework. The exercise states: Let $T_0 = \inf\{s > 0 : B_s = 0\}$ and let $R = \inf\{t > 1 : B_t = 0\}$. Use the Markov property at time 1 to get $$P_x(R>1+t) = \int_{-\infty}^\infty p_1(x,y)P_y(T_0>t)dy.$$ Here, $P_x$ is the Wiener measure associated to Brownian motion started at $x$ and I think $p_1(x,y) = \frac{1}{\sqrt{2\pi t}}e^{-(x-y)^2/2t}$ is the density of Brownian motion started at $x$ but I'm not sure.
I cannot even see where to begin. I don't see how $P_y(T_0>t)$ will be involved. I don't see what function $Y$ to apply the Markov property to. Some explanation of how to approach this would be much appreciated.
Consider $\tilde B_s:=B_{s+1}$, $s\ge 0$, and notice that $R-1=\inf\{s>0:\tilde B_s=0\}=T_0(\tilde B)$. Now condition on where $B$ is at time $1$: the density of $B_1$ is $y\mapsto p_1(x,y)$ (as you surmise) and the conditional distribution of $\tilde B$, given that $B_1=y$, is $P_y$. Putting these thoughts together, $$ \eqalign{ P_x(R>1+t) &=\int_{-\infty}^\infty p_1(x,y)P_x(T_0(\tilde B)>t|B_1=y)\,dy\cr &=\int_{-\infty}^\infty p_1(x,y)P_y(T_0>t)\,dy.\cr } $$
More detail on the above calculation. Using the shift operator, $\{R>1+t\}=\theta_1^{-1}(\{T_0>t\})$. The Markov property (8.2.2) tells us that $$ \eqalign{ P_x(R>1+t; B_1\in C) &=P_x(\theta_1^{-1}\{T_0>t\}; B_1\in C)\cr &=E_x[\varphi(B_1); B_1\in C]\cr &=\int_C \varphi(y)p_1(x,y)\,dy, } $$ where $\varphi(y)=P_y(T_0>t)$, and $C$ is any Borel subset of $\Bbb R$. In particular, taking $C=\Bbb R$ we get $$P_x(R>1+t) =\int_{-\infty}^\infty p_1(x,y)P_y(T_0>t)\,dy. $$