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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 $n-1$?

Since $T_n=1+2+ \dots+n = \frac{n(n+1)}{2}$ I have that $4T_n=\frac{4n(n+1)}{2}=2n(n+1)$

So I am to compute $\gcd(2n(n+1),n-1)$? Using long division I got that $$\frac{2 n^{2} + 2 n}{n - 1}=2 n + 4+\frac{4}{n - 1} $$ but I don't know how to proceed?

Jiina Jojlo
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    So, what you wrote shows that that $2n(n+1)=(2n+4)(n-1)+4$. thus the gcd in question must be a divisor of $4$. Is $4$ achievable? – lulu May 26 '22 at 20:53
  • By Euclidean reduction $(n-1,f(n))=(n-1,f(1)),$ as explained here in the dupe, where $,f(x),$ is any polynomial with integer coef's. So $,(n-1,2n(n+1)) = (n-1,4),$ is a divisor of $4$, which is at most $4$ (and equals $4\iff 4\mid n-1,,$ i.e. $n = 1+4k,$ for some integer $k)\ \ $ – Bill Dubuque May 27 '22 at 08:30

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It's trivial to prove that gcd$(n-1,n+1) = 1 $ or $2$ depending on their parity.

Since, we want the greatest possible value of the gcd, we will assume that gcd$(n-1,n+1)= 2$ (It means both of them are even numbers) . Clearly, gcd$(2n(n+1),n-1) \ge 2$.Since,gcd$(n, n-1) = 1$, $n$ will not have any effect on gcd$(2n(n+1),n-1)$. Now, observe that in every pair of two consecutive even numbers, one of the numbers will be a multiple of $4$.If $n-1$ is a multiple of $2$ , the gcd will remain $2$. On the contrary, if $(n-1)$ is multiple of $4$, then $(n+1)$ will only be divisible by $2$ , therefore $2n(n+1)$ will also be divisible by $4$. So, finally we got our result, gcd$(2n(n+1), n-1) = 4$.

VVR
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