Given 5 integers show that you can find two whose sum or difference is divisible by 6.

HINT: If two of the numbers have the same remainder on division by $6$, their difference is divisible by $6$, so you can focus on the case in which all five integers have different remainders on division by $6$. Try to show that no matter which of the $6$ possible remainders is missing, you must have two numbers whose sum is a multiple of $6$.

Simple.

As noted above if you have two different integers which give same remainder when divided by 6, then their difference will be divisible by 6.

Consider case of integers which all have diffrent reminders when divided by 6. There can only be 6 remainders then - 0,1,2,3,4,5 It is easy to notice that 1+5 = 6 which means that sum integers which give remainder of 1 and 5 when divided by 6 will be divisible by 6, and 2+4 = 6, which means that sum integers which give remainder of 2 and 4 when divided by 6 will be divisible by 6,

Since you need to choose 5 integers from the six possible types of remainders, you cannot elude both pairs. Therefore at least one sum will be divisible by 6 QED

$a\equiv b\pmod{\! m}$ denotes '$a,b$ leave same remainders when divided by $m$', or $m\mid a-b$ (see Modular Arithmetic).

$x^2\equiv \{0,1,3,4\}\pmod{\! 6}$ for any $x\in\Bbb Z$ ($\color{#0af}4$ possible remainders).

To see why, square $6k,\, 6k\pm 1,\, 6k\pm 2,\, 6k+ 3$.

Let your integers be $a_1,a_2,a_3,a_4,a_5$ ($\color{#0af}{4}+1$ integers).

By pigeonhole principle exist $i\neq j$ with $a_i^2\equiv a_j^2\pmod{\! 6}$, or i.e. $6\mid (a_i+a_j)(a_i-a_j)$

Assume for contr. that $6\nmid a_i+a_j, a_i-a_j$. Then WLOG $3\mid a_i+ a_j,\, 2\mid a_i- a_j$

But $a_i+ a_j=a_i- a_j+ 2a_j$, so $a_i+ a_j$ is even too and $6\mid a_i+ a_j$, contr assumption.

This used elementary methods, but it generalizes. $\,x^2\not\equiv 2$, $\,x^2\not\equiv -1$$\pmod{\! 6}$ is a consequence of quadratic reciprocity, because $3$ (a prime divisor of $6$) is of the form $8k+3$. So you did not need to square $0,\pm 1,\pm 2, 3$, which is neat since the $6$ could've been larger.