Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$

Question

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

Answer 1

$$ \frac{n^2(n^2-1)(n^2+1)}{60} = 2n\binom{n+2}5 + 2\binom{n+1}4 + \binom{n+1}3. $$

Answer 2

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 $n^2-1\equiv 0\mod 4$.

Answer 3

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 $1 \pmod 5$.

Answer 4

Note that we have $$n^2(n^2-1)(n^2+1)=n^2(n^2-1)(n^2-4)+5n^2(n^2-1)$$ and that $n^2$ is divisible by $4$ or $n^2-1$ is divisible by $8$.

Answer 5

Note that $n^5-n=n(n^2-1)(n^2+1)$ is divisible by $5$ by little Fermat. $n^3-n$ is likewise divisible by $3$ and either $n^2$ or $n^2-1$ is divisible by $4$.