Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now enough to prove that $n(n^2+1)$ is divisible by $10$. But for $n=4$, $4(17)\ne10\lambda$ but for $n=4$, $\text{P}$ is $4080=60\cdot68$ which means apart from just being a multiple of $6$, $n(n-1)(n+1)$ is actually helping $n(n^2+1)$ with a $5$ to sustain divisibility by $60$.
How to tackle this? Thanks

Use the Chinese Remainder theorem to show this product is congruent to $0\bmod 3, 4$ and $5$. You'll determine first what the squares  are, modulo these numbers.
A sketch for the case of the modulus $4$:
Every number $n\equiv 0,1,2$ or $3\: (\equiv -1)\bmod 4$. So 
$$n^2\equiv 0^2=0,\: 1^2=1, 2^2\equiv 0\quad\text{or}\quad 3^2\equiv(-1)^2=1.$$
As a conclusion, either $n^2$  or $n^2-1\equiv 0\mod 4$.
