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  "body": "\n## Proof\n\n1. Let $n$ be a [[composite number|0N3ADXHNTQ]]. Then $n \u003e 1$ and $n$ is not a [[prime number|0SN0MAEKYX]].\n2. So $n$ has a positive [[divisor|096PW7Q014]] $a$ with $a \\neq 1$ and $a \\neq n$. Write $n = a b$ with an integer $b$.\n3. Since $n \u003e 0$ and $a \u003e 0$, $b \\ge 1$, so $a \\le n$. Hence $1 \u003c a \u003c n$.\n4. Then $b = n / a$ satisfies $1 \u003c b \u003c n$. If $b = 1$ then $a = n$, and if $b = n$ then $a = 1$; both are excluded.\n5. By swapping $a$ and $b$ if needed, we may assume $a \\le b$. Then $a^2 \\le a b = n$.\n6. By the [[existence of a prime divisor|0SPBEDVNW1]], since $a \\ge 2$, there is a prime $p$ with $p \\mid a$.\n7. Since $p$ is a positive divisor of the positive integer $a$, $p \\le a$. So $p^2 \\le a^2 \\le n$.\n8. Since $p \\mid a$ and $a \\mid n$, $p \\mid n$. So $p$ is a prime factor of $n$ with $p^2 \\le n$, that is, $p \\le \\sqrt{n}$.\n\n## Idea\n\nA composite number splits as $n = ab$ with two factors larger than $1$. The smaller factor is at most $\\sqrt{n}$, since otherwise the product would exceed $n$. Any prime factor of this smaller factor works.\n",
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