Leicich wrote: Determine all positive integers $n$ such that the number $n(n+2)(n+4)$ has at most $15$ positive divisors. Using the fact that the number of positive divisors is a multiplicative function, it appears that only possible values are 1, 2, 3, 4, 5, 7, 9

kprepaf wrote: Leicich wrote: Determine all positive integers $n$ such that the number $n(n+2)(n+4)$ has at most $15$ positive divisors. Using the fact that the number of positive divisors is a multiplicative function, it appears that only possible values are 1, 2, 3, 4, 5, 7, 9 Can anyone please elaborate in some details?

My Solution: Denote the number of divisors of a number $n$ by $\tau (n)$. It is known that $\tau$ is an arithmetic function. So, $\tau (mn) = \tau (m) \tau (n)$, whenever $(m,n) = 1$. We take two cases: $\bullet$ $n$ is odd: We see that $n, n+2, n+4$ are pairwise co prime. So, $\tau (n (n+2)(n+4)) = \tau (n) \tau (n+2) \tau (n+4)$. Given that $\tau (n (n+2)(n+4)) \leq 15$, we get $\tau (n) \tau (n+2) \tau (n+4) \leq 15$. When $n=1$ this trivially holds. Else, we have $\tau (n) \ge 2$. So, the only possibilities of $\tau (n), \tau (n+2), \tau (n+4)$ are from the set $\{ 2,3 \}$. There is at most one $3$ among $\tau (n), \tau (n+2), \tau (n+4)$. When one of them is a $3$, then it must be a square of a prime. When one of them is a $2$, then the number is prime. When $\tau (n) = 3$, we have $n=p^2$ for some prime $p$. Also, we must have $n+2 = p^2 + 2$ to be prime, and taking modulo $3$, we find that $n=9$ is the only such number. This clearly satisfies the equation. Now, similarly, modulo $3$ shows that if one of $n, n+2, n+4$ is a square of a prime and the other two are primes, then $n=9, n=7, n=5$ are the only solutions. When all the three are primes, we have $n=3$ to be the only solution, taking modulo $3$ again. The only odd $n$ satisfying the conditions are $\boxed{1,3,5,7,9}$. $\bullet$ $n$ is even: Let $n=2k$ for some positive integer $k$. We have $n(n+2)(n+4) = 8k(k+1)(k+2)$. Now we see that $k,k+1,k+2$ are pairwise coprime, and thus, proceeding similar to the previous case, we must have $k=1,2$, which gives the only even solutions of $n$ to be $\boxed{2,4}$, which indeed satisfy the conditions. So, $$\boxed{n \in \{ 1,2,3,4,5,7,9\}}$$ Q E D Alternatively, a solution not involving the $\tau$ function is also possible.

I analyze first when $n \ge 11.$ We treat $2$ cases Case 1. If $n$ is odd. Then $gcd(n, n + 2) = gcd(n, n + 4) = gcd(n + 2, n + 4 ) = 1$ and with certainty one of them is divisible by $3$ so we have at least $4$ prime factors who divide these product so at least $16$ divisors. Case 2. $n$ is even. Then i write $n = 2k$ and we have $n (n + 2) (n + 4) = 8 \cdot k \cdot (k + 1) \cdot (k + 2).$ At least one of $k,k+1,k+2$ is even. So we have $2^4$ divide the product. If $k$ is odd then $gcd(k,k+2) = 1$ so we have at least two prime numbers,resulting in at least $5 \cdot 2 \cdot 2 = 20$ divisor for the product. If $k$ is even then we have $2^6 \mid n(n + 2) \cdot (n + 4)$ and we have at least two prime odd number so we have at least $7 \cdot 2 \cdot 2 = 28$ divisors. So we haven't solution for $n \ge 11.$ After calculation, we find out that the values of $n$ that satisfy the condition are: \[ n \in \{ 1,2,3,4,5,7,9 \} \]