# Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Let a be an arbitrary positive integer, then corresponding to the positive integers a and 6, there exist, non-negative integers q and r such that

a = 6q + r, where 0 ≤ r < 6

Squaring both the sides using (a + b)2 = a2 + 2ab + b2

a2 = (6q + r)2 = 36q2 + r2 + 12qr

a2 = 6(6q2 + 2qr) + r2

where,0 ≤ r < 6

Case I When r = 0 we get

a2 = 6(6q2) = 6m

where, m = 6q2 is an integer.

Case II when r = 1 we get

a2 = 6(6q2 + 2q) + 1 = 6m + 1

where, m = (6q2 + 2q) is an integer.

Case III When r = 2 we get

a2 = 6(6q2 + 4q) + 4 = 6m + 4

where, m = (6q2 + 4q) is an integer.

Case IV When r = 3 we get

a2 = 6(6q2 + 6q) + 9

a2 = 6(6q2 + 6a) + 6 + 3

a2 = 6(6q2 + 6q + 1) + 3 = 6m + 3

where, m = (6q + 6q + 1) is an integer.

Case V when r = 4 we get

a2 = 6(6q2 + 8q) + 16

a2 = 6(6q2 + 10q) + 24 + 1

a2 = 6(6q2 + 8q + 2) + 4 = 6m + 4

where, m = (6q2 + 8q + 2) is an integer

Case VI When r = 5 we get

a2 = 6(6q2 + 10q) + 25

a2 = 6(6q2 + 10q) + 24 + 1

a2 = 6(6q2 + 10q + 4) + 1 = 6m + 1

where, m = (6q2 + 10q + 1) + 1 is an integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

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