What is the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$

Question

I want to find the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$

from $n(n-1)(n+1)$ 2 and 3 should divide this expression for all positive n. how can I find the rest? which python says $(2, 3, 6, 7, 9, 42, 14, 18, 21, 63,126)$

Answer 1

If $$n\equiv 0 \mod 3,$$ then the factor $$n^2+3\equiv 0 \mod 3.$$ Otherwise $$n^2\equiv1 \mod 3$$ and thus $$n^2+5\equiv 0\mod3.$$ Therefore $$(n^2+3)(n^2+5)\equiv0\mod3.$$ Let us now look modulo $7$ and compute the number for all elements of $\mathbb Z_7$ (which means to check if the number is a multiple of $7$ for $n=0,\dots 6$. You can easily check that it is true. Therefore this number is a multiple of $2\times3\times3\times7=126$ and all its divisors. There are no others since for $n=2$, $n(n^2-1)(n^2+3)(n^2+5)=126\times3$ but for $n=3$, $n(n^2-1)(n^2+3)(n^2+5)=126\times 32$.