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Based on my current knowledge, the formula for finding the product of all positive divisors of a number $n$ is $= n^{\frac{\tau (n)}{2}}$ where the function $\tau (n)$ outputs the number of positive divisors for that number $n$. However, wouldn't this formula only work for those and only those numbers who have an even number of divisors? In that case, what is the method used to determine the product of all positive divisors of a number with an odd number of positive divisors? Thank you in advance.

Clyde Kertzer
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Camelot823
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    If $n$ has an odd number of positive divisors, $n$ is a perfect square! – ETS1331 Apr 19 '23 at 18:07
  • It works for all numbers. Did you try an example? – Karl Apr 19 '23 at 18:12
  • See this post - it works. – Dietrich Burde Apr 19 '23 at 18:14
  • I don't see how it works. Say, $100$; the prime factorization of which is $2^2\times 5^2$. It's positive divisors are ${1, 2, 4, 5, 10, 20, 25, 50, 100}$. This sequence has the property that the first and last, second and penultimate, third and third from end ... all produce the number $100$ when multiplied. So if we pair these numbers, and form $k$ pairs, then the product of all positive divisors of $100$ would equal $k\times 100$. However, if we have an odd number of positive divisors, then one term in the sequence will be left unpaired. So how exactly does the formula work then? – Camelot823 Apr 19 '23 at 19:03
  • Maybe, the missing information is that $n^{\frac p2}= \sqrt{n^p}$ – Lourrran Apr 19 '23 at 19:04
  • I perfectly understand the formula, I simply don't see how it was derived for all cases – Camelot823 Apr 19 '23 at 19:04
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    This answer by André Nicolas explains both cases perfectly: https://math.stackexchange.com/a/1126019/ Therefore, I am closing this as a duplicate. – Mike Earnest Apr 19 '23 at 21:34
  • @MikeEarnest ahh... Thank you very much! This is exactly what I needed! – Camelot823 Apr 19 '23 at 22:00

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