Question

For each ineteger a, 3 does not divide a in and only if 3 divides a^2+1

Answer 1

Answer:

Show that if a is an integer, then 3 divides a3−a

we can write, where k is an integer;

a3−a=3k

a(a2−1)=3k

Now if a=k then

a2−1=3 and a=±2 so a3−a=24=3×8

If a is not equal to k;

then

a(a2−1)=a(a+1)(a−1)=3k

since a(a+1)(a−1) is the product of 3 consecutive integers, the expression is divisible by 3.

Is this ok?, just a check. I'm not really all that good at number theory.