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.

Hint: There is a formula for $T_n$ in terms of n. – D.B. Mar 28 '19 at 17:20 — D.B., Mar 28 '19 at 17:20
Just plug in some values and you will see a pattern. – Infiaria Mar 28 '19 at 17:21 — Infiaria, Mar 28 '19 at 17:21

score 2 · Accepted Answer · answered Mar 28 '19 at 17:51

2

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

answered Mar 28 '19 at 17:51

lhf

216,483

score 2 · Answer 2 · answered Mar 29 '19 at 01:55

2

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 $\,(4T_n,n\!-\!1)= (4n+\color{#c00}{4T_{n-1}},\,n\!-\!1) = (4n,n\!-\!1) = (4,n\!-\!1)$

answered Mar 29 '19 at 01:55

Bill Dubuque

272,048

score 1 · Answer 3 · answered Mar 28 '19 at 20:10

1

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

answered Mar 28 '19 at 20:10

Triangular numbers and gcd

3 Answers3