On division by 6, a² cannot leave remainder (a N)

From Euclid’s Division lemma, for any positive integer aN.

a = 6m + r where, 0<r<6 and m N

a = 6m, a = 6m + 1, a = 6m + 2, a = 6m + 3, a = 6m + 4 or a = 6m + 5

Now, a = 6m

Then a² = 6²m²

= 36m² = 6 × (6m)² + 0

Here, remainder is 0.

If a = 6m + 1

Then a² = (6m + 1)²

= 36m² + 12m + 1

= 6(6m² + 2m) + 1

Here, remainder is 1.

If a = 6m + 2

Then a² = (6m + 2)²

= 36m² + 24m + 4

= 6(6m² + 4m) + 4

Here, remainder is 4.

If a = 6m + 3

Then a² = (6m + 3)²

= 36m² + 32m + 9

= 6(6m² + 6m + 1) + 3

Here, remainder is 3.

If a = 6m + 4

Then a² = (6m + 4)²

= 36m² + 48m + 16

= 6(6m² + 8m + 2) + 4

Here, remainder is 4.

If a = 6m + 5

Then a² = (6m + 5)²

= 36m² + 60m + 25

= 6(6m² + 10m + 4) + 1

Here, remainder is 1 = .

Thus in any case, if a² is divided by 6, then the remainder is 0, 1, 3 or 4 but not 5.