Prove that for any integer $a$ we have $6 \mid (a^3 + 5a)$.
We'll use the Division Algorithm which guarantees that given integers $a$ and (divisor) $b \geq 1$, there exists unique integers quotient $q$ and remainder $r$ such that $a$ can be written as
$$a = qb + r \quad \text{where} \quad 0 \le r < b$$
If we choose $b=6$, then $0 \le r < 6$, and we have
$$ a^3 + 5a = 216 q^3 + 108 q^2 r + 18 q r^2 + 30 q + r^3 + 5 r$$
The integer $r$ can only be 0,1,2,3,4,5. Let's consider each case:
- $r=0$ means $ a^3 + 5a = 216 q^3 + 30 q = 6(36 q^3 + 5q)$, divisible by 6.
- $r=1$ means $ a^3 + 5a = 216 q^3 + 108 q^2 + 18 q + 30 q + 6 = 6(36 q^3 + 18q^2 + 8q + 1)$, divisible by 6.
- $r=2$ means $ a^3 + 5a = 216 q^3 + 216 q^2 + 102 q + 18 = 6(36q^3 + 36q^2 + 17q+3)$, divisible by 6.
- $r=3$ means $a^3 + 5a = 6 (36 q^3 + 54 q^2 + 32 q + 7)$, divisible by 6.
- $r = 4$ means $a^3 + 5a = 6 (36 q^3 + 72 q^2 + 53 q + 14)$, divisible by 6.
- $r = 5$ means $a^3 + 5a = 6 (36 q^3 + 90 q^2 + 80 q + 25)$, divisible by 6.
So, for any integer $a$ we have $6 \mid (a^3 + 5a)$.