Q. 44.1( 25 Votes )

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

Answer :

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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