For each integer a if a is an odd integer that is not a mul

• For each integer a, if a is an odd integer that is not a multiple of 3, then a 2 1 (mod 6).

Solution

For each integer a, if a is an odd integer that is not a multiple of 3, then a is of the form 6k ± 1.

So a^2 = 36k^2 ± 12k + 1 = 12k(3k ± 1) +1

case 1: k even, then 12k is a multiple of 24 so a ^2 is divisibly a multiple of 6, plus 1.

case 2: k odd, then (3k ± 1) is even,so we have a^2 is of the form 12k*even +1,which is also divisible by multiple of 6, plus 1.

Thus in both the cases a ^2 is divisibly a multiple of 6, plus 1 implies that a ^2 - 1 is divisibly a multiple of 6

which shows that a² 1 ( mod 6 )