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  "body": "\n## Proof\n\nWe prove existence for $n = 2, 4, p^k, 2p^k$, and then nonexistence for every other $n \\ge 2$. Throughout, $\\varphi$ is [[Euler's totient function|0VV04M41AN]] and $\\operatorname{ord}_n(x)$ is the [[multiplicative order|0SBRX5CZ5R]] of an integer $x$ [[coprime|0GFKP8F8AG]] to $n$. A [[primitive root|0KM2CJ80G7]] modulo $n$ is such an $x$ with $\\operatorname{ord}_n(x) = \\varphi(n)$.\n\n1. Let $e = \\operatorname{ord}_n(x)$ and let $m \\ge 0$ with $x^m \\equiv 1 \\pmod n$. By the [[division algorithm|0CHQKZFK5S]], $m = qe + r$ with $q \\ge 0$ and $0 \\le r \u003c e$. Then $1 \\equiv x^m = (x^e)^q x^r \\equiv x^r \\pmod n$, using [[powers of congruences|0J5VGJYYC3]]. Minimality of $e$ forces $r = 0$. So $e \\mid m$.\n2. By [[Euler's theorem|0VC4DVE7DQ]], $x^{\\varphi(n)} \\equiv 1 \\pmod n$. So $\\operatorname{ord}_n(x) \\mid \\varphi(n)$ by step 1, and in particular $\\operatorname{ord}_n(x) \\le \\varphi(n)$.\n\nExistence.\n\n3. For $n = 2$: $\\varphi(2) = 1$, and $x = 1$ has order $1$. For $n = 4$: $\\varphi(4) = 2$, and $3 \\not\\equiv 1$ while $3^2 = 9 \\equiv 1 \\pmod 4$. So $3$ has order $2$.\n4. Let $p$ be an odd prime and $k \\ge 1$. By the [[existence of primitive roots modulo a prime|04XGKK8WQW]], there is $g$ with $\\operatorname{ord}_p(g) = p - 1$. Then $p \\nmid g$.\n5. Suppose $g^{p-1} \\equiv 1 \\pmod{p^2}$. By the [[binomial theorem|0EHWPWGHB2]], $(g + p)^{p-1} = g^{p-1} + (p-1) g^{p-2} p + p^2 s$ for an integer $s$.\n6. So $(g+p)^{p-1} \\equiv 1 + p^2 g^{p-2} - p g^{p-2} \\equiv 1 - p g^{p-2} \\pmod{p^2}$. By [[Euclid's lemma|051VYKA1EX]], $p \\nmid g^{p-2}$, so $p^2 \\nmid p g^{p-2}$. Hence $(g+p)^{p-1} \\not\\equiv 1 \\pmod{p^2}$.\n7. Also $g + p \\equiv g \\pmod p$, so all powers agree modulo $p$ and $\\operatorname{ord}_p(g + p) = p - 1$. Replacing $g$ by $g + p$ if needed, we may assume $g^{p-1} \\not\\equiv 1 \\pmod{p^2}$.\n8. By [[Fermat's little theorem|0M25PRN3WQ]], $g^{p-1} \\equiv 1 \\pmod p$. So $g^{p-1} = 1 + p t_1$ with $t_1$ an integer, and $p \\nmid t_1$ by step 7.\n9. For $1 \\le i \\le p - 1$, we have $p \\mid \\binom{p}{i}$. Indeed, $\\binom{p}{i} \\, i! \\, (p-i)! = p!$ by the [[factorial formula|033KTKKRV6]]. Here $p \\mid p!$, but $p \\nmid i! \\, (p-i)!$ by Euclid's lemma, since all factors are less than $p$. So Euclid's lemma gives $p \\mid \\binom{p}{i}$.\n10. Claim: for each $j \\ge 1$, $g^{p^{j-1}(p-1)} = 1 + p^j t_j$ with $t_j$ an integer and $p \\nmid t_j$. We argue by induction on $j$. Step 8 is the case $j = 1$.\n11. Assume the claim for $j$ and write $t = t_j$. By the binomial theorem, $g^{p^j (p-1)} = (1 + p^j t)^p = 1 + p^{j+1} t + \\sum_{i=2}^{p-1} \\binom{p}{i} p^{ij} t^i + p^{pj} t^p$.\n12. For $2 \\le i \\le p - 1$, the term is divisible by $p \\cdot p^{2j}$ by step 9, and $2j + 1 \\ge j + 2$. The last term is divisible by $p^{pj}$, and $pj \\ge 3j \\ge j + 2$.\n13. So $g^{p^j(p-1)} = 1 + p^{j+1}(t + p u)$ for an integer $u$. Set $t_{j+1} = t + pu$. Since $p \\nmid t$, also $p \\nmid t_{j+1}$. This proves the claim.\n14. Now let $e = \\operatorname{ord}_{p^k}(g)$, which is defined since $p \\nmid g$. By step 2 and the [[totient of a prime power|0AVY8HKBGD]], $e \\mid p^{k-1}(p-1)$.\n15. From $g^e \\equiv 1 \\pmod{p^k}$ we get $g^e \\equiv 1 \\pmod p$. By step 1 with modulus $p$, $(p - 1) \\mid e$. Write $e = (p-1) c$. Then $c \\mid p^{k-1}$.\n16. By the [[fundamental theorem of arithmetic|084HH1NDCB]], $c = p^i$ with $0 \\le i \\le k - 1$.\n17. Suppose $i \\le k - 2$; this needs $k \\ge 2$. Then $e \\mid p^{k-2}(p-1)$, so $g^{p^{k-2}(p-1)} \\equiv 1 \\pmod{p^k}$.\n18. But step 10 with $j = k - 1$ gives $g^{p^{k-2}(p-1)} = 1 + p^{k-1} t_{k-1}$ with $p \\nmid t_{k-1}$. So $p^k$ does not divide $g^{p^{k-2}(p-1)} - 1$, a contradiction.\n19. Hence $i = k - 1$ and $e = p^{k-1}(p-1) = \\varphi(p^k)$. So $g$ is a primitive root modulo $p^k$.\n20. Let $n = 2p^k$. Put $h = g$ if $g$ is odd, and $h = g + p^k$ if $g$ is even. Since $p^k$ is odd, $h$ is odd.\n21. As $h \\equiv g \\pmod{p^k}$, all powers of $h$ and $g$ agree modulo $p^k$. So $\\operatorname{ord}_{p^k}(h) = \\varphi(p^k)$.\n22. A prime that divides both $h$ and $2p^k$ is $2$ or $p$. But $h$ is odd and $p \\nmid h$. So $h$ is coprime to $2p^k$.\n23. By the [[multiplicativity of the totient|0FSJA9M43W]] and the [[totient of a prime|0HY1MXMEN9]], $\\varphi(2p^k) = \\varphi(2)\\varphi(p^k) = \\varphi(p^k)$.\n24. Let $e = \\operatorname{ord}_{2p^k}(h)$. From $h^e \\equiv 1 \\pmod{p^k}$ and step 1, $\\varphi(p^k) \\mid e$, so $e \\ge \\varphi(p^k)$. By step 2, $e \\le \\varphi(2p^k) = \\varphi(p^k)$. So $h$ is a primitive root modulo $2p^k$.\n\nNonexistence.\n\n25. Let $k \\ge 3$. Claim: every odd $a$ satisfies $a^{2^{k-2}} \\equiv 1 \\pmod{2^k}$. We argue by induction on $k$.\n26. For $k = 3$, write $a = 2b + 1$. Then $a^2 = 4b(b+1) + 1$. One of $b$ and $b + 1$ is even, so $8 \\mid a^2 - 1$.\n27. If $a^{2^{k-2}} = 1 + 2^k s$ with $s$ an integer, then $a^{2^{k-1}} = 1 + 2^{k+1} s + 2^{2k} s^2$. Since $2k \\ge k + 1$, this is $1$ modulo $2^{k+1}$. This proves the claim.\n28. The integers coprime to $2^k$ are the odd ones. By step 27 and step 1, each has order at most $2^{k-2}$. This is less than $2^{k-1} = \\varphi(2^k)$, so there is no primitive root modulo $2^k$.\n29. Claim: $\\varphi(c)$ is even for every integer $c \\ge 3$. Let $S$ be the set of $x$ with $1 \\le x \\le c$ and $\\gcd(x, c) = 1$. Then $c \\notin S$.\n30. For $x \\in S$, we have $1 \\le c - x \\le c - 1$, and $\\gcd(c - x, c) = \\gcd(x, c) = 1$, since $x$ and $c - x$ have the same common divisors with $c$. So $x \\mapsto c - x$ maps $S$ to itself, and applying it twice gives $x$.\n31. If $x = c - x$, then $c = 2x$ and $\\gcd(x, c) = x$, so $x = 1$ and $c = 2$. This is excluded. So $S$ splits into pairs $\\set{x, c - x}$, and $\\varphi(c) = \\abs{S}$ is even.\n32. Claim: if $n = ab$ with $a, b \\ge 3$ coprime, then there is no primitive root modulo $n$. Let $x$ be coprime to $n$. Then $x$ is coprime to $a$ and to $b$.\n33. Let $L = \\varphi(a)\\varphi(b)/2$. It is an integer, and $L = \\varphi(a) \\cdot (\\varphi(b)/2) = \\varphi(b) \\cdot (\\varphi(a)/2)$ with integer factors, by step 31.\n34. By Euler's theorem, $x^L = (x^{\\varphi(a)})^{\\varphi(b)/2} \\equiv 1 \\pmod a$ and $x^L = (x^{\\varphi(b)})^{\\varphi(a)/2} \\equiv 1 \\pmod b$.\n35. Both $x^L$ and $1$ are congruent to $1$ modulo $a$ and modulo $b$. By the [[Chinese remainder theorem|0FNN9B4W64]], $x^L \\equiv 1 \\pmod{ab}$.\n36. So $\\operatorname{ord}_n(x) \\le L \u003c \\varphi(a)\\varphi(b) = \\varphi(n)$, by multiplicativity of the totient. This proves the claim.\n37. Now let $n \\ge 2$ not be of the form $2, 4, p^k, 2p^k$. Write $n = 2^j m$ with $j \\ge 0$ and $m$ odd. If $m = 1$, then $n = 2^j$ with $j \\ge 3$, and step 28 applies.\n38. If $m$ has two distinct prime factors, let $q$ be one of them and $q^f$ the exact power of $q$ in $m$. Then $m = q^f m'$ with $q \\nmid m'$. The other prime factor of $m$ divides $m'$, so $m'$ is odd and $m' \\ge 3$. Take $a = q^f$ and $b = 2^j m'$.\n39. Then $a, b \\ge 3$, and $q \\nmid b$ by Euclid's lemma. So $a$ and $b$ are coprime, and step 36 applies.\n40. If $m = q^f$ for one odd prime $q$, then $j \\ne 0$ and $j \\ne 1$ by assumption. So $j \\ge 2$. Take $a = 2^j \\ge 4$ and $b = q^f \\ge 3$, which are coprime. Step 36 applies.\n41. So a primitive root modulo $n$ exists if and only if $n \\in \\set{2, 4, p^k, 2p^k}$.\n\n## Idea\n\nA primitive root modulo $p$ lifts to $p^k$ once $g^{p-1}$ is not $1$ modulo $p^2$. Then each $p$-th power raises the exact power of $p$ in $g^{p^{j-1}(p-1)} - 1$ by one, so no smaller exponent works. For other moduli the group splits into two parts of even size, or is a large power of $2$, and a common exponent below $\\varphi(n)$ kills every unit.\n",
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