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  "body": "\n## Proof\n\nWe take $a$ as large as possible.\n\n1. Let $n$ be a positive integer. Let $A$ be the set of positive integers $a$ such that $a^2$ [[divides|096PW7Q014]] $n$.\n2. The set $A$ is nonempty, because $1^2 = 1$ divides $n$.\n3. The set $A$ is bounded. If $a \\in A$, then $a^2$ is a positive divisor of the positive integer $n$, so $a \\le a^2 \\le n$.\n4. So $A$ is a nonempty finite set, and it has a largest element $a$.\n5. Write $n = a^2 b$ with an integer $b$. Since $n \u003e 0$ and $a^2 \u003e 0$, $b$ is a positive integer.\n6. Suppose that $b$ is not [[square-free|004JW7A2N5]]. Then there is a [[prime number|0SN0MAEKYX]] $p$ with $p^2 \\mid b$. Write $b = p^2 c$ with an integer $c$.\n7. Then $n = a^2 p^2 c = (a p)^2 c$, so $a p \\in A$.\n8. Since $p \\ge 2$, $a p \u003e a$. This contradicts the choice of $a$ as the largest element of $A$.\n9. So $b$ is square-free, and $n = a^2 b$ with positive integers $a$ and $b$.\n\n## Idea\n\nTake the largest square that divides $n$ and divide it out. If the quotient still had a factor $p^2$, a larger square would divide $n$. So the quotient is square-free.\n",
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        "locator": "Square-free factors of integers",
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  "name": "Proof that every positive integer is a square times a square-free integer",
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