Let $n$ be a positive integer, and consider its prime factorization. For every prime $p$ that appears in $n$ an odd number of times (meaning, an odd exponent), we put it in exactly once. For every prime that appears in $n$ an even number of times, we don't put it in. We then multiply all the primes that we put in, and call the resulting number $m$. I wonder, is $m$ the smallest positive integer such that if you multiply it by $n$, you get a perfect square?
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It seems to be true. – M.Riyan Aug 08 '24 at 10:10
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As here in the linked dupe, by existence and uniqueness of prime factorizations, $,n,$ is a square $\iff$ every prime factor of $,n,$ occurs to even power. So scaling $,n,$ to a square requires that we include at least one factor of $,p,$ for each prime occurring to odd power, so the least scale factor $,m,$ includes exactly one of each. – Bill Dubuque Aug 08 '24 at 17:09
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Consider the number $n$ and its prime factors be $a,b,c,d$ [Suppose]. The prime numbers appear respectively $2p,2q,2r-1,2s-1$ times and so we can write: $$n=a^{2p}b^{2q}c^{2r-1}d^{2s-1}$$ Taking only the odd exponents of prime, we make the number $m=cd$, where we have taken them only once. Multiply $n$ and $m$, we get: $$nm=a^{2p}b^{2q}c^{2r-1}d^{2s-1} \cdot cd=a^{2p}b^{2q}c^{2r}d^{2s}$$ And $mn$ is surely a perfect square, since $\sqrt{mn}=a^pb^qc^rd^s$.
M.Riyan
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