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

The product of three consecutive numbers is divisible by $3$.
If $n$ is even, then $4\mid n^2$. Otherwise $n^2-1$ and $n^2 + 1$ are both even.
Now the hard part:
\begin{array}{c|c}
n \pmod 5 & n^2 \pmod 5\\ \hline
0 & 0 \\
1 & 1 \\
2 & 4 \\
3 & 4 \\
4 & 1
\end{array}
or, without modular notation,
\begin{array}{c|c}
n & n^2 & \text{The remainder when dividing by five}\\ \hline
5k & 25k^2 & 5(5k)\\
5k+1 & 25k^2 + 10k + 1 & 5(5k^2 + 2k) + 1 \\
5k+2 & 25k^2 + 20k + 4 & 5(5k^2 + 4k) + 4 \\
5k+3 & 25k^2 + 30k + 9 & 5(5k^2 + 6k + 1) + 4 \\
5k+4 & 25k^2 + 40k + 16 & 5(5k^2 + 8k + 3) + 1 \\
\end{array}
No matter what $n$ is, $n^2$ is either $-1, 0$, or $1 \pmod 5$.
