Prove that $n(n+1)(n+2)$ where $n\in N$ and $n+1=k^2$ ($k\in N$) is divisible by $60$. My Attempt
Let $N=n(n+1)(n+2)$
Now $n=k^2-1 \implies N=(k^2-1)k^2(k^2+1)$
Now if $k=2m \implies N=(4m^2-1)4m^2(4m^2+1)$ which is divisible by $3$ and $4 \implies$ It is divisible by $12$.
Now if $k=2m+1 \implies N=(4m^2+4m)(4m^2+4m+1)(4m^2+4m+2)$ which is divisible by $8$.
This is as far as I got. How do I prove its divisible by $60$? Sorry if the formatting isn't the best.

To conclude you have to show that it is divisible by $5$ so you have to see what the possible remainders of $n+1$ when dividing by $5$ are, and conclude that at least one of $n,n+1$ and $n+2$ is a multiple of $5$.

 If $n+1$ is a square then $n+1\equiv -1,0$ or $1\bmod 5$. It follows at least one among $n,n+1,n+2$ is $0 \bmod 5$ as desired.

