if m is an integer show that 3 divides (m^{3 }- m)
$\displaystyle m^3 - m \ = $
$\displaystyle m(m^2 - 1) \ = $
$\displaystyle m(m - 1)(m + 1) \ = $
$\displaystyle (m - 1)m(m + 1)$
This is a product of three consecutive integers.
Consecutive multiples of three differ by three, and one of (m - 1), m, and (m + 1) will be a multiple of three.
So, that product will be divisible by 3.
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But if you wanted to, you could make a stronger statement.
Consecutive multiples of two differ by two, and either one of (m - 1), m, and (m + 1) will be a multiple of two,
or two of (m - 1), m, and (m + 1) will be a multiple of two, depending on the value of m.
So, the product (m - 1)m(m + 1) will also be divisible by 2.
Then the stronger statement can be:
"If m is an integer, then 6 divides $\displaystyle \ m^3 - m $."