Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.

First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $2$ that isn't the multiple of $4$. Since the problem is asking for a minimal case that satisfies these conditions, $n+1$ should be the multiple of $4$, $n+2$ should be the multiple of $3$, and $n+3$ should be the multiple of $2$, thus giving $4 \cdot 3^2 \cdot 2^3 = 288.$
However, I am unsure as to whether or not this answer is correct, as testing cases has consistently yielded higher GCDs than this.
 A: Following the comment of lulu, look at numbers of the form $n=19+24k$.
For $k=0$, $$(n+1)(n+2)^2(n+3)^3(n+4)^4=20\cdot 21^2\cdot 22^3\cdot 23^4$$ which has the factorization $2^5\cdot 3^2\cdot 5\cdot 7^2\cdot 11^3\cdot 23^4$ and is not divisible by any number containing more than five factors of $2$ and two factors of $3$
For $k=1$, the product $44\cdot 45^2 \cdot 46^3\cdot 47^4$ has no factors of $7$, so $7$ cannot be a factor of the divisor you seek.
For $k=2$, the product $68\cdot 69^2 \cdot 70^3\cdot 71^4$ has no factors of $11$, so $11$ cannot be a factor of the divisor you seek.
For $k=4$, the product $116\cdot 117^2 \cdot 118^3\cdot 119^4$ has no factors of $5$ or $23$, so neither $5$ nor $23$ can be a factor of the divisor you seek.
So the only factor of $20\cdot 21^2\cdot 22^3\cdot 23^4$ that can and must appear in the divisor you seek is $2^53^2$.
It's hard to see that in comparing two examples at a time, as many of them will have a $5$ or a $7$, as well as the occasional larger prime factor in common.
A: Let $f(x) \in \Bbb Z[x]$. We have $m \mid f(n) \iff f(n)\equiv 0\pmod m$, but when written as this it's clear that $n$ may be replaced by any integer (not only natural numbers) congruent to $n$ modulo $m$, to get a equivalent congruence. The means that
"The largest positive integer that divide $f(n)$ for all positive integers $n$"
is the same as
"The largest positive integer that divide $f(n)$ for all integers $n$"
Now to this particular case:
We have
$$f(-5) =(-5+1)(-5+2)^2(-5+3)^3(-5+4)^4 = (-4)(-3)^2(-2)^3(-1)^4=2^5 3^2$$
(so the answer can't be larger than that) and you argued that $f(n)$ is always multiple of $2^53^2$ so that's the answer.
A: First write $f(n)=(n+1)(n+1)^2(n+2)^3(n+4)^4$ The gcd of the $f(n)$s; $n \in \mathbb{N}$, is of the form $2^b3^a$, for some positive integers $a$ and $b$. Indeed, no prime $p \ge 7$ divides $f(1)$, and $5$ does not divide $f(5)$. Thus indeed, the  common prime divisors are restricted to $2$ and $3$, and thus the gcd of the $f(n)$s is indeed of the form $2^b3^a$.
We now find $b$. Observe that the largest power of $2$ that divides $f(3)$ is $5$. Meanwhile, the largest power of $3$ that divides $f(1)$ is $2$. So $a \le 2$ and $b \le 5$. As you have shown $a \ge 2$ and $b \ge 5$, it follows that the equations $a=2$ and $b=5$ hold. This leaves us with the gcd of the $f(n)$s; $n \in \mathbb{N}$, is indeed $2^53^2$.
