Let $d_n= \text{lcm} \{1,2,3..., n \}$ then for which natural numbers $n$ is $$\frac{24 d_n^5}{(n+1)^5} \text{not an integer?}$$

We know that $$ \text{lcm}\{1,2,3,...,n\}.\text{gcd}\{1,2,3,...,n\}=n!$$ So we have $$d_n=\frac{n!}{\text{gcd}\{1,2,3,...,n\}}$$ Hence $$\frac{24 d_n^5}{(n+1)^5}=24\left(\frac{n!}{(n+1)\text{gcd}\{1,2,3,...,n\}}\right)^5$$ Please help me with this problem. Thank you!

  • 2
    $\begingroup$ If you take $n=5$ you have $d_5=60$ that is divisible for $n+1=6$ and so their fifth powers are divisible as well $\quad$ In general I think that the quantity you describe is not an integer iif $n=p-1$ for some prime $\endgroup$
    – Marco
    Commented Jun 11 at 11:10
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    $\begingroup$ $\text{lcm}\{1,2,3,...,n\}.\text{gcd}\{1,2,3,...,n\}=n!$ is incorrect. In fact, $\text{gcd}\{1,2,3,\ldots,n\}=1$ and $\text{lcm}\{1,2,3,...,n\}\neq n!$ for many $n$'s. $\endgroup$ Commented Jun 11 at 11:43
  • $\begingroup$ @SungjinKim Thanks. For which $n$ is my number not an integer? $\endgroup$
    – Max
    Commented Jun 11 at 11:48
  • $\begingroup$ The question is unclear. Do you mean "for all integers $n$, the quantity is not an integer" (which is false, as whenever $n+1$ is not a prime power, it divides $d_n$) or "it is not true that for all integers $n$ the quantity is an integer" (which is true)? $\endgroup$ Commented Jun 11 at 11:52
  • $\begingroup$ @MassimilianoFoschi My question is that for which type of integers $n$, the quantity is not an integer. $\endgroup$
    – Max
    Commented Jun 11 at 11:54

1 Answer 1


As mentioned by Massimiliano, if $n+1$ is not a prime power, then $24d_n^5/(n+1)^5$ is an integer. Let $n+1=dk$ where $1<d\leq n, 1<k\leq n$ and $(d,k)=1$. We have $d|d_n$ and $k|d_n$. This gives $dk|d_n$. Then $24d_n^5/(n+1)^5=24d^5k^5 m / d^5k^5$ for some positive integer $m$. Hence, $24d_n^5/(n+1)^5=24m$ is an integer.

On the other hand, if $n+1=p^k$ is a prime power, then $p^{k-1} || d_n$. This gives $p^{5(k-1)}||d_n^5$ and $p^{5k}||(n+1)^5$. For this prime $p$, the denominator has $p^5$ after making $d_n^5/(n+1)^5$ in the lowest terms. However, $24$ does not have $5$-th powers in the factorization. Hence, we conclude that

$24d_n^5/(n+1)^5$ is not an integer if and only if $n+1$ is a prime power.

  • $\begingroup$ $\,q\in\Bbb Q,\ 24q^5\in\Bbb Z\!\iff\! q\in\Bbb Z\,$ (else $1<d^5\mid 24\,$ for $\,d=$ reduced denom of $q)\,$. So it reduces to when $\,n\!+\!1\nmid {\rm lcm}(1,2,\ldots,n),\,$ easily seen to be true $\!\iff\! n\!+\!1$ is a prime power. Both of these inferences are duplicates. $\ \ $ $\endgroup$ Commented Jun 11 at 17:54
  • $\begingroup$ ($+1$) Please prove that if $n+1$ is not a prime power, then $24d_n^5/(n+1)^5$ is an integer. $\endgroup$
    – Max
    Commented Jun 11 at 19:35
  • $\begingroup$ @Max It is proved in the first paragraph. $\endgroup$ Commented Jun 11 at 20:33

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