MATHmaniac aka CrUNchy

prove that for any integer a, 3|a^3-a

let a be an integer.when we divide by 3,
we have 3 possiblities:


a=3k
a=3k+1
a=3k+2

we want to show that 3 divides a^3-a
we know that a^3=a(a^2-1)=a(a-1)(a+1)

1st case,
if a=3k,then a^3=3.k(3k-1)(3k+1)
Thus,3|a^3-a for a=3k

2nd case,
if a=3k+1,Then a^3-a=3((3k+2)(3k+1)(k))
Thus 3|a^3-a for a=3k+1

3rd case,
if a=3k+2,then a^3-a=3((3k+2)(3k+1)(k+1))
Thus, 3|a^3-a for a=3k+2


since 3|a^3-a for all cases,then we can conclude that 3|a^3-a for any integer a

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