Question

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

Answer 1

Step-by-step explanation:

let a be any positive odd integer and b= 4

By Euclids division algorithm

a= bq+r

Therefore,a= 4q +r

where,0<=r<=4

Therefore r= 0,1,2,3,4

a=4q or 4q+1 or 4q+2 or 4q+3

here, a is any positive odd integer so we write a= 4q+1 or 4q+3 where q is some integer.

Answer 2

Answer:

Step-by-step explanation:

An even number is always in the form of 2n

Since it is divisible by 2 and

An odd number is always in the form of 2n+1

or 2n-1 as it can come just one before or one after the even number (here n is any whole number)

Let's proceed

Taking odd number as the form of 2n+1

If we square this we get

(2n+1)² = 4n²+4n+1

=4(n²+n)+1

=4q+1

We can also solve it by taking odd no. as 2n-1