Question

show that the square of any odd integer is of the form 4 Q + 1 for some integer q

Answer 1

By Euclid’s division algorithm, we have a = bq + r, where 0 On putting b = 4 in Eq. (i), we get
a = 4q + r, where 0 ≤ r < 4 i.e., r = 0,1,2,3 ...(ii)
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 q, (4q+1) and (4q+3) are odd integers.
Now, a2 = (4q+1)2 = 16q2 + 1 + 8q = 4(4q2 + 2q) + 1 [∵(a+b)2 = a2 + 2ab + b2]
is a square which is of the form 4m+ 1, where m = (4q2 + 2q) is an integer.
and a2 = (4q + 3)2 = 16q2 + 9 + 24q = 4(4q2 + 6q + 2) + 1 is a square [∵(a+b)2 = a2 + 2ab + b2]
which is of the form 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.

Answer 2

odd number =2n+1
(2n+1)^2=4(n^2+n)+1=4q+1