# Question on LCM and divisibility

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!

• 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 Commented Jun 11 at 11:10
• $\text{lcm}\{1,2,3,...,n\}.\text{gcd}\{1,2,3,...,n\}=n!$ 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. Commented Jun 11 at 11:43
• @SungjinKim Thanks. For which $n$ is my number not an integer?
– Max
Commented Jun 11 at 11:48
• 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)? Commented Jun 11 at 11:52
• @MassimilianoFoschi My question is that for which type of integers $n$, the quantity is not an integer.
– Max
Commented Jun 11 at 11:54

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 $$1 and $$(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.

• $\,q\in\Bbb Q,\ 24q^5\in\Bbb Z\!\iff\! q\in\Bbb Z\,$ (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. $\ \$ Commented Jun 11 at 17:54
• ($+1$) Please prove that if $n+1$ is not a prime power, then $24d_n^5/(n+1)^5$ is an integer.
– Max
Commented Jun 11 at 19:35
• @Max It is proved in the first paragraph. Commented Jun 11 at 20:33