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This may seem like a weird question, but how to find the numbers with non-exponential prime factors i.e. numbers which specifically have primes that occur only once in its prime-factorization. For eg: $15=3×5$, $70=2×5×7$. So, 15 and 70 are such numbers which have prime-factors that occur only once in its prime-factorization.

$28=7×4$. In its prime-factorization = $7×2×2$ So, 28 is NOT a number which has prime-factors that occur only once.

So, is there any way to find such numbers??

Thank you.

Dietrich Burde
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Ishan.J
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  • "non-exponential" is bad terminology. You should say repeated prime factors. They are called square-free numbers. You can get them by multiplying different prime factors (ex:235711*13). – Eric Apr 03 '21 at 17:37
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    The official term is "square-free" (as none of their factors [except $1$] can be a square). What do you mean "how do you find them"? You multiply together one or more distinct primes and ... there they are.... – fleablood Apr 03 '21 at 17:45
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    I really have no idea what you mean "how to find them". You can make a list of them. If $p_i$ are all the primes in order then we can make exhaustive systematic list. $2,3,2\cdot 3,5,2\cdot 5,3\cdot 5, 2\cdot 3\cdot 5, 7, 2\cdot 7... etc$. We can describe this as every time we come to a new prime multiply that prime by all the terms on our list so for..... but serious what on earth do you mean how do you find them? They are right there! What's to be found? – fleablood Apr 03 '21 at 17:51
  • You can write every positive integer in a prime factorization as $\prod p_i^{k_i}$ where $p_i$ is a prime and $k_i$ is a positive integer. So $28 = 2^2 \cdot 7$. The square free one are those where all the powers are just one. $14 = 2\times 7$ is on. Given any number such as $50,400=2^53^25^27$ and "squish it" to $2\cdot 3\cdot 5\cdot7 =210$. And if you have a number and want to see if it square free.... well, just prime factor it and look if it is. – fleablood Apr 03 '21 at 18:00
  • @DietrichBurde in your edit, are you sure the OP was asking about how to detect if a given number were square-free rather than how to find a square-free number? – fleablood Apr 03 '21 at 18:07
  • @fleablood To be honest, I am not sure. I thought this would make more sense, since writing down square free numbers is "trivial". – Dietrich Burde Apr 03 '21 at 18:28
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    There is no efficient algorithm known to decide whether a given number is squarefree. – Peter Apr 04 '21 at 11:57

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There is an algorithm for "detecting square free numbers". See for example the article by Andrew R. Booker, Ghaith A. Hiary, Jon P. Keating:

"We present an algorithm, based on the explicit formula for $L$-functions and conditional on the generalized Riemann hypothesis, for proving that a given integer is squarefree with little or no knowledge of its factorization. We analyze the algorithm both theoretically and practically and use it to prove that several RSA challenge numbers are not squarefull."

Dietrich Burde
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