Triangular numbers and gcd

Question

For all positive integers $n$, the $n$th triangular number $T_n$ is defined as $T_n = 1+2+3+ \cdots + n$. What is the greatest possible value of the greatest common divisor of $4T_n$ and $n-1$?

Can someone help me to start this? I have no idea what to do.

Answer 1

Hint: Euclidean division gives $2n(n+1)=(2n+4)(n-1)+4$.

Answer 2

$2((n-1)+1)((n-1)+2)=2(n-1)^2+6(n-1)+4$ shows at most 4 can be their gcd.

Answer 3

Can also be done via (additive) Wilson theorem, i.e. pairing up additive inverses mod $\,n\!-\!1\,$ shows $\,\color{#c00}{2T_{n-1}}\equiv 0\pmod{\!n\!-\!1},\,$ so $\,(4T_n,n\!-\!1)= (4n+\color{#c00}{4T_{n-1}},\,n\!-\!1) = (4n,n\!-\!1) = (4,n\!-\!1)$