A hyper cube $Q_n$ is a graph that have the length-n binary sequences as its vertices. Two vertices are adjacent if they differ in one entry.
I found a Hamilton cycle for $Q_3$ as follows $$000 \to 100 \to 110 \to 010 \to 011 \to 111 \to 101 \to 001 \to 000$$
Now I want to use this cycle to find a Hamiltonian cycle for $Q_4$, I notice that we can split the vertices of $Q_4$ into two groups those that ends with $0$ and those that ends with $1$. If we remove the ending zero from the first group, we are left with the vertices in $Q_3$ and if remove the ending $1$ from the second group we are also left with the vertices in $Q_3$.
I though about just joining the hamiltion cycle that I did for $Q_3$ twice, once where we have the ending is zero and once where we have the ending is $1$
$$000\color{red}{0} \to 100\color{red}{0} \to 110\color{red}{0} \to 010\color{red}{0} \to 011\color{red}{0} \to 111\color{red}{0} \to 101\color{red}{0} \to 001\color{red}{0}$$
and when we have the ending is $1$
$$000\color{blue}{1} \to 100\color{blue}{1} \to 110\color{blue}{1} \to 010\color{blue}{1} \to 011\color{blue}{1} \to 111\color{blue}{1} \to 101\color{blue}{1} \to 001\color{blue}{1}$$
So now If I join them
$$000\color{red}{0} \to 100\color{red}{0} \to 110\color{red}{0} \to 010\color{red}{0} \to 011\color{red}{0} \to 111\color{red}{0} \to 101\color{red}{0} \to \color{green}{001\color{red}{0} \to 000\color{blue}{1}} \to 100\color{blue}{1} \to 110\color{blue}{1} \to 010\color{blue}{1} \to 011\color{blue}{1} \to 111\color{blue}{1} \to \color{green}{101\color{blue}{1} \to 001\color{blue}{1}}$$
There is a problem, you it doesn't differ by $1$ any more when I join them (See the green coloured ones) so how could I fix this to construct a valid hamiltion cycle for $Q_4$