Question 5
Show that the square of any odd integer is of the form 4m + 1, for some integer m.
By Euclid's division algorithm, we have a=bq+r, where 0≤r<b
On putting b = 4, we get,
a=4q+r where 0≤r<4 i.e., r = 0, 1, 2, 3
If r = 0 ⇒ a = 4q is divisible by 2 ⇒ 4q is even.
If r = 2⇒ a=4q+2,2(2q+1) is divisible b y 2 ⇒ 2 (2q + 1) is even.
If r = 3⇒ a=4q+3,(4q+1) and (4q+3) are odd integers.
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]
This 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.