Question

Show that square root of any odd integer is of the form of 4m+1 for some integer m ?

Answer 1

please refer the attachment.
Hope this helps you quickly

Answer 2

Hi Mate !!
Hi Mate !!


Let a be any square root of odd integer which when divided by 4 gives m as qoutient and r as remainder.

By Euclid's Division lemma

a = bq + r

0 ≤

Let a be any square root of odd integer which when divided by 4 gives q as qoutient and r as remainder.

By Euclid's Division lemma

a = bq + r

a = 4m + r
0 ≤ r < b

Here b is 4
So, r = 0 , 1 , 2 , 3

• a = 4m

• a = 4m + 1

• a = 4m + 2

• a = 4m + 3

So, 4m + 1 is an odd number because 1 is odd number. [ though 3 is also odd one but we haven't to show 4m + 3 ]