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  "body": "\n## Statement\n\nLet $n$ be an integer with $n \u003e 1$.\nIf $(n - 1)! \\equiv -1 \\pmod n$, then $n$ is a [[prime number|0SN0MAEKYX]].\n\n## Remarks\n\nWith [[Wilson's theorem|03KGGKTB2E]], this gives: an integer $n \u003e 1$ is prime if and only if $(n - 1)! \\equiv -1 \\pmod n$.\nThe proof: if $n = a b$ with $1 \u003c a \u003c n$, then $a$ divides $(n - 1)!$. A congruence $(n - 1)! \\equiv -1 \\pmod n$ would give $a \\mid 1$, which is false.\nIn fact $(n - 1)! \\equiv 0 \\pmod n$ for each composite $n \u003e 4$, and $3! \\equiv 2 \\pmod 4$.\nThis primality test is not practical, because $(n - 1)!$ needs about $n$ multiplications.\n\n## Explanations\n\n### L1 human\n\nFor a composite number $n$, a factor of $n$ appears in the product $1 \\cdot 2 \\cdots (n - 1)$.\nSo the product cannot leave the remainder $n - 1$. For example, $5! = 120 \\equiv 0 \\pmod 6$.\n",
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  "name": "Converse of Wilson's theorem",
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