Question

prove that square of positive integer is in form 4m or 4m +1

Answer 1

Answer:

Step-by-step explanation:

Let "A" be any positive integer and "b=4".

Apply edl for this.

Possible remainders are 0,1,2 or 3.

Therefore A can be in the form of

4q+1,4q+2,4q+3

Now, (4q)2=4(4)*Q2

=16q2

=4(4q2)

=4m

Again,(4q+1)2=(4q)2+1+8q

=16q2+8q+1

=4(4q2+2q)+1

=4m+1

Therefore square of all positive integers is of the form 4m,4m+1.

Answer 2

Let a be any +ve integer b=2 then by euclids division lemma a=bq+r such that 0 lessthan or equal to r less than b therefore the possible remainders are 0,1

a=(2q)^2=4q^2=4m (where m is some +ve integer)

a=(2q+1)^2 = 4q^2+4q+1=4m+1 (where m is some +ve integer)

Therefore the square of any +ve integer is of the form 4m or 4m+1

Hence proved