Show that for each $2<p<q<k$, $Q_k$ has a cycle of length $2^p+2^q$.

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Let $Q_k$ be the $k$-hypercube. Assume that $Q_k$ has a Hamiltian cycle for each $k \geq 2$. Assume that $ 4 \leq k$. Show that for each $ 2 \leq p < q < k$, $Q_k$ has a cycle of length $2^p + 2^q$.

I am not sure how to prove this. Please help.

I am thinking of using induction on the number of vertices.

Note: $Q_k$ can be decomposed into 2 copies of hypercubes in lower dimension => $Q_{k-1} Q_{k-1}$.

Note: Hamiltonian cycle is a Hamiltonian path that starts and ends at the same vertex, which visits each vertex exactly once.

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Every hypercube is Hamiltonian. (You can prove this easily by induction on the dimension of the hypercube.) Now let $e_1, \ldots, e_k$ be unit coordinate vectors (i.e. $e_i$ has a $1$ in the $i$th coordinate slot and $0$ everywhere else) and consider the disjoint sub-hypercubes $$S_p = \{c_1 e_1 + \ldots + c_p e_p | (\forall 1 \leq i \leq p) c_i \in \{0, 1\} \}$$ and $$S_q = \{c_1 e_1 + \ldots + c_q e_q + e_k | (\forall 1 \leq i \leq q) c_i \in \{0, 1\} \}$$

Take Hamiltonian paths on $S_p$ and $S_q$ such that the path on $S_p$ joins the origin to $e_1$ and such that the path on $S_q$ joins $e_k$ to $e_1 + e_k$. (Once you have one Hamiltonian path on a hypercube, you can rotate it to connect the origin to whichever adjacent vertices you choose.) Then splice the two paths together by severing the connections $0 \to e_1$ in $S_p$ and $e_k \to e_1 + e_k$ in $S_q$ and instead connecting $0 \to e_k$ and $e_1 \to e_1 + e_k$. This gives you the necessary cycle.