jmc

Question

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

Accepted Answer

By the arithmetic series formula, T_n = n(n+1)2, so 4T_n = 2n(n+1) = 2n^2 + 2n. By the Euclidean algorithm, align*gcd,(2n^2 + 2n, n-1) &= gcd,(2n^2 + 2n - (n-1) 2n, n-1) &= gcd,(4n, n - 1) &= gcd,(4n - 4(n-1) , n-1) &= gcd,(4, n -1) 4.align* For example, this is true for n = 5.