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A colleague of mine mentioned that there was something special about the number nine. He noticed that nine had three positive integer divisors {1, 3, 9} and the square root of nine is three.

His question was: Are there any other positive integers n, such that the number of positive integer divisors of n is equal to the square root of n?

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2 Answers 2

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Yes, $1$ has positive integer divisor $\{1\}$, and $|\{1\}|=1=\sqrt{1}$.

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  • $\begingroup$ Yes, one works! Any more numbers with the given property? $\endgroup$ Jul 20, 2022 at 21:27
  • $\begingroup$ Almost sure 1 and 3 are the only solutions. Type Length[Divisors[n^2]]-n, n=1...1000 into Wolfram Alpha, only 1st and 3rd equal zero and after 30 seems no possibility to achieve it again. wolframalpha.com/… $\endgroup$
    – z100
    Jul 20, 2022 at 21:55
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The number of positive divisors is a multiplicative function in the number theoretical sense. The square root is also multiplicative. We can therefore proceed prime factor by prime factor. If p occurs in n 2k times (an even number because n is a perfect square) then this contributes a factor of p^k to the square root and a factor of 2k+1 to the divisor count function. This rules out p=2. But for anything other than p=3,k=1 p^k will be greater than 2k+1. This leaves 1 and 9 as the only possible solutions.

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    $\begingroup$ This should be the accepted answer. $\endgroup$
    – RobPratt
    Jul 21, 2022 at 13:08
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    $\begingroup$ @RobPratt I debated whether I should have accepted your answer or Loopy’s. Loopy’s answer is more thorough than yours but I eventually decided to accept yours because 1) your answer was posted first AND 2) your answer technically answers my question the way I wrote it AND 3) I found your answer concise and easier to understand. $\endgroup$ Jul 21, 2022 at 19:24
  • $\begingroup$ OK, thanks @WillOctagonGibson! $\endgroup$
    – RobPratt
    Jul 21, 2022 at 20:13

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