Question

show that the square of any positive integer cannot be of the form 6 m + 2 or 6 m + 5 for any Integer m​

Answer 1

Let a be an arbitrary positive integer, then by Euclid’s division algorithm, 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

a = 6q + r, where 0 ≤ r < 6

⇒ a2 = (6q + r)2 = 36q2 + r2 + 12qr [∵(a+b)2 = a2 + 2ab + b2]

⇒ a2 = 6(6q2 + 2qr) + r2 ...(i)

where,0 ≤ r < 6

Case I When r = 0, then putting r = 0 in Eq.(i), we get

a2 = 6 (6q2) = 6m

where, m = 6q2 is an integer.

Case II When r = 1, then putting r = 1 in Eq.(i), we get

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

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

Case III When r = 2, then putting r = 2 in Eq(i), we get

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

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

Case IV When r = 3,then putting r = 3 in Eq.(i), we get

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

= 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, then putting r = 4 in Eq.(i) we get

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

= 6(6q2 + 8q) + 12 + 4

⇒ a2 = 6(6q2 + 8q + 2) + 4 = 6m + 4

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

Case VI When r = 5, then putting r = 5 in Eq.(i), we get

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

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

⇒ a2 = 6(6q2 + 10q + 4) + 1 = 6m + 1

where, m = (6q2 + 10q + 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.

Answer 2

Answer:

Step-by-step explanation:

Let the square of any posititve integer be in the form 6q + 2, 6q +5

x = 6q +2

Squaring both sides

Xsquare = 36(q)square + 4 + 24q

x square = 36(q) square - 6+ 2 + 24q

xsquare = 6 [ 6(q) square - 1 + 4q] + 2

put [ 6(q) square - 1 + 4q] = m

Therefore x = 6m + 2