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  "body": "\n## Proof\n\nWe prove both directions. Step 1 is a fact about closed walks that both directions use.\n\n1. Let $(z_0, z_1, \\dots, z_m)$ be a closed walk with $m \\ge 3$, $z_m = z_0$, and $z_1, \\dots, z_m$ pairwise distinct. Then its $m$ edges are pairwise distinct, so it is a [[circuit|0PD57PSHNR]], and hence a [[cycle|07CT9RDY7Q]].\n2. Proof of step 1. Read the indices modulo $m$. The edges are $\\set{z_{t-1}, z_t}$ for $t = 1, \\dots, m$. Two of them can be equal only if the pairs share both vertices.\n3. Since $z_1, \\dots, z_m$ are distinct, the equality $\\set{z_{s-1}, z_s} = \\set{z_{t-1}, z_t}$ with $s \\ne t$ forces $z_{s-1} = z_t$ and $z_s = z_{t-1}$. Then $t \\equiv s - 1$ and $s \\equiv t - 1$, so $2 \\equiv 0$ modulo $m$, which is false for $m \\ge 3$.\n4. ($\\Rightarrow$) Let $G$ be a [[tree|0MJ0JS72DZ]]. It is [[connected|0FFAYNMAJZ]], so it has a vertex and any two vertices $u, v$ are joined by a [[walk|0VJZM0F9YM]].\n5. By [[every walk contains a path|09VZBHB33V]], there is a [[path|0JDE56HWQ8]] from $u$ to $v$.\n6. Suppose $P = (p_0, \\dots, p_a)$ and $Q = (q_0, \\dots, q_b)$ are different paths from $u$ to $v$.\n7. Neither is a proper initial part of the other. For example, if $a \u003c b$ and $p_t = q_t$ for $t \\le a$, then $q_a = p_a = v = q_b$, which contradicts the distinct vertices of $Q$.\n8. So there is a least index $i$ with $1 \\le i \\le \\min(a, b)$ and $p_i \\ne q_i$. Then $p_{i-1} = q_{i-1}$; call this vertex $x$.\n9. Let $j \\ge i$ be the least index such that $p_j = q_l$ for some $l \\ge i$. It exists, because $p_a = v = q_b$ and $b \\ge i$.\n10. Consider the closed walk $C = (x, p_i, p_{i+1}, \\dots, p_j, q_{l-1}, q_{l-2}, \\dots, q_i, x)$, where $p_j = q_l$ and $q_{i-1} = x$. Its length is $(j - i + 1) + (l - i + 1)$.\n11. The vertices $p_i, \\dots, p_j$ are distinct and differ from $x = p_{i-1}$, since $P$ is a path. The vertices $q_{l-1}, \\dots, q_i$ are distinct and differ from $x = q_{i-1}$, since $Q$ is a path.\n12. No $q_t$ with $i \\le t \u003c l$ equals some $p_s$ with $i \\le s \\le j$. For $s \u003c j$ this follows from the minimality of $j$, and for $s = j$ from $p_j = q_l$ and $t \\ne l$.\n13. So the vertices of $C$ after the first are pairwise distinct. If the length were $2$, then $j = l = i$ and $p_i = q_i$, which is false. So the length is at least $3$.\n14. By step 1, $C$ is a cycle in $G$. This contradicts that $G$ is [[acyclic|0XNN9Z8E5A]]. Hence the path from $u$ to $v$ is unique.\n15. ($\\Leftarrow$) Suppose $G$ has a vertex, and any two vertices are joined by exactly one path.\n16. A path is a walk, so any two vertices are [[reachable|0J8XE7W68F]] from each other. So $G$ is connected.\n17. Suppose $G$ has a cycle $(w_0, w_1, \\dots, w_k)$. Its length $k$ is at least $3$: $k = 1$ needs a loop, and $k = 2$ uses the edge $w_0 w_1$ twice.\n18. Then $(w_0, w_1)$ and $(w_k, w_{k-1}, \\dots, w_1)$ are paths from $w_0 = w_k$ to $w_1$. The second has distinct vertices because $w_1, \\dots, w_k$ are distinct.\n19. These two paths have lengths $1$ and $k - 1 \\ge 2$, so they are different. This contradicts uniqueness, so $G$ is acyclic.\n20. By steps 16 and 19, $G$ is a tree.\n\n## Idea\n\nTwo different routes between the same two vertices must split apart at some vertex and later meet again. The two pieces between the split and the first meeting point form a cycle. Conversely, a cycle gives two different routes between neighbouring vertices on it.\n",
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