Show that the square of any integer is of the form 4m + 1, for some integer m.


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


Here, b is any positive integer .


On putting b = 4 we get


a = 4q + r, where 0 ≤ r < 4 (i)


If r = 0


a = 4q, 4q is divisible by 2


4q is even.


If r = 1


a = 4q + 1, (4q + 1) is not divisible by 2.


If r = 2


a = 4q + 2,2(2q + 1) is divisible by 2


2(2q + 1) is even.


If r = 3


a = 4q + 3, (4q + 3) is not divisible by 2.


So, for any positive integer (4q + 1) and (4q + 3) are odd integers.


Case I For (4q + 1)


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


a2 = 16q2 + 1 + 8q


a2 = 4(4q2 + 2q) + 1 = 4m + 1


Where m = (4q2 + 2q) in an integer.


Case II For (4q + 3)


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


a2 = 16q2 + 9 + 24q = 16q2 + 24q + 8 + 1


a2 = 4 (4q2 + 6q + 2) + 1 = 4m + 1


where m = (4q2 + 6q + 2) is an integer.


Hence, for some integer m, the square of any odd integer is of the form 4m + 1.


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