If \(n\) is an integer, prove the following is also an integer:
$$\dfrac{n(n + 1)}{2}$$

We'll prove \(2 \mid n(n + 1)\) both when \(n = 2k\) and when \(n = 2k + 1.\) For the first case, \(n(n + 1) = 2k(2k + 1)\) is obviously divisible by \(2.\) For the second case, \(n(n + 1) = (2k + 1)(2k + 1 + 1) = (2k + 1)(2k + 2),\) which is divisible by \(2\) because \((2k + 2) = 2(k + 1).\)

If \(n\) is an integer, prove the following is also an integer:
$$\dfrac{n(n + 1)(n + 2)}{6}$$

Case 1. \(6k(6k + 1)(6k + 2)\) is divisible by \(6.\)

Case 2. \((6k + 1)(6k + 2)(6k + 3)\) is divisible by \(2\) and \(3.\)

Case 3. \((6k + 2)(6k + 3)(6k + 4)\) is divisible by \(2\) and \(3.\)

Case 4. \((6k + 3)(6k + 4)(6k + 5)\) is divisible by \(2\) and \(3.\)

Case 5. \((6k + 4)(6k + 5)(6k + 6)\) is divisible by \(6.\)

Case 6. \((6k + 5)(6k + 6)(6k + 7)\) is divisible by \(6.\)

Let \(n \in \mathbb{Z}\). Then \(n \le n^2\)

Let \(n \in \mathbb{Z}\). Then \(n^2 + 3n + 2\) is even.

Let \(x, y \in \mathbb{Z}\) such that both \(xy\) and \(x + y\) are even. Then both \(x\) and \(y\) are even.

Let \(n \in \mathbb{Z}\). Then \(3n^2 + n + 14\) is even.

Let \(n \in \mathbb{Z}\). Then \(x^2 + 3x + 1\) is odd.

Let \(n \in \mathbb{Z}\). Then \(n^2 + 3n + 5\) is odd.

\(\max(x, y) + \min(x, y) = x + y\)

For all integers \(n\), if \(5 \nmid n\), then \(n^2 \equiv 1 \pmod{5}\) or \(n^2 \equiv 4 \pmod{5}\).

For all \(x \in \mathbb{R}\), \(\lfloor x \rfloor + \lfloor -x \rfloor = 0\) or \(-1\).

If \(2 \le n \le 5\), then \(4 \nmid n^2 + 2\).

Let \(n \in \mathbb{Z}\). Then \(4 \nmid n^2 + 2\).

A perfect cube is of the form \(9k - 1\), \(9k\), or \(9k + 1\), for some integer \(k\).

A perfect square is of the form \(3k\) or \(3k + 1\), for some integer \(k\).

Let \(x, y \in \mathbb{Z}\). Then \(x\) and \(y\) have the same parity if and only if \(x + y\) is even.

Prove:
$$\max(x, y) = \dfrac{1}{2}(x + y) + \dfrac{1}{2}\lvert x - y \rvert$$