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  "body": "\n## Statement\n\nLet $n$ be a positive integer.\nThen there are positive integers $a$ and $b$ such that $n = a^2 b$ and $b$ is [[square-free|004JW7A2N5]].\n\n## Remarks\n\nThe integers $a$ and $b$ are unique. The number $b$ is the product of the primes $p$ for which the exponent of $p$ in $n$ is odd.\nFor example, $72 = 2^3 \\cdot 3^2 = 6^2 \\cdot 2$, so $a = 6$ and $b = 2$.\nThe proof uses the [[prime factorization|0AEKDFCYFA]] of $n$. Write each exponent $e$ as $e = 2 q + r$ with $r \\in \\{0, 1\\}$.\n\n## Explanations\n\n### L1 human\n\nTake out of $n$ all the pairs of equal primes that you can. The pairs give the square $a^2$.\nAt most one copy of each prime stays, so the rest $b$ is square-free.\nFor example, $50 = 2 \\cdot 5 \\cdot 5 = 5^2 \\cdot 2$.\n",
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