For how many integers \(n\), where \(2 \le n \le 2026\), is \(\dfrac{(n-1)(n)(n+1)}{8}\) equal to an integer?
We can solve this problem by considering whether \(n\) is even or odd.
If \(n\) is an odd integer, then \((n-1)\) and \((n+1)\) are both even. In fact, \((n-1)\) and \((n+1)\) are consecutive even integers, so one is a multiple of \(4\) and the other is divisible by \(2\) (since it is even). Thus, \((n-1)(n+1)\) is divisible by \(4 \times 2 = 8\). Therefore, for any odd integer \(n\), \(\dfrac{(n-1)(n)(n+1)}{8}\) is equal to an integer. Since \(2 \le n \le 2026\), there are \(1012\) possible odd values for \(n\).
If \(n\) is an even integer, then \((n-1)\) and \((n+1)\) are both odd. Thus, \((n-1)(n)(n+1)\) is divisible by \(8\) only when \(n\) is divisible by \(8\). Since \(2 \le n \le 2026\), the largest value of \(n\) that is a multiple of \(8\) is \(2024\). Since \(8 \times 253=2024\), it follows that there are \(253\) multiples of \(8\) between \(2\) and \(2026\).
Therefore, there are \(1012+253=1265\) integers \(n\), with \(2 \le n \le 2026\), such that \(\dfrac{(n-1)(n)(n+1)}{8}\) is equal to an integer.