The two conditions $$\frac{p_{\mathrm{up}}(n)}{p_{\mathrm{down}}(n+1)}=c \quad \text{and} \quad p_{\mathrm{up}}(n)+p_{\mathrm{down}}(n)=1$$
lead to $p_{\mathrm{up}}=\frac{c}{c+1}$ and $p_{\mathrm{down}}= \frac{1}{c+1}$.
Also $p_{\mathrm{up}},p_{\mathrm{down}}$ are probabilities therefore $p_{\mathrm{up}},p_{\mathrm{down}},c\in [0,1]$.
One could proof the forms for $p_{\mathrm{up}},p_{\mathrm{down}}$ easily by induction, but I'd like a constructive proof much better.
However I failed to come up with one. Are there multiple solutions or am I just to bad to find the construction?
For $c=0$, we get $p_{\mathrm{up}}(n)=0$ and $p_{\mathrm{down}}(n)=1$ for all $n$; which is of the $\left(\frac{c}{c+1},\frac{1}{c+1}\right)$ kind.
For $c>0$, the most general form of a solution looks as follows: $$\begin{eqnarray} (-c)^n\left(p_{\mathrm{up}}(n) - \frac{c}{c+1}\right) & = &\left(p_{\mathrm{up}}(0) - \frac{c}{c+1}\right) \\ (-c)^n\left(\frac{1}{c+1} - p_{\mathrm{down}}(n)\right) & = & \left(\frac{1}{c+1} - p_{\mathrm{down}}(0)\right) \end{eqnarray}$$
This makes the constant solution $\left(\frac{c}{c+1},\frac{1}{c+1}\right)$ obvious; $n$ doesn't play any role in such case. If $0 < c < 1$, the term $(-c)^n$ goes to zero, so the other one, $\left(p_{\mathrm{up}}(n)-\frac{c}{c+1}\right)$, must (in absolute value) go to infinity; violating the permitted range of $[0,1]$ for $p_{\mathrm{up}}(n)$. Thus, there are no other solutions.
The case of $c=1$ is slightly more interesting: we get $p_{\mathrm{down}}(n+1)=p_{\mathrm{up}}(n)$ (and vice versa), so the sequence just alternates up- and down-values indefinitely.