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.