Question

Show that the square of any positive odd integer, is of the form 4m+1 for some integer m.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

Let a be any odd positive integer ,then on divided a by b, we have

a= bq+r,    0≤ r<b ………(1)

[By Euclid division Lemma]

On putting b= 2 in equation (1) we get

a=2q+r,           0≤ r<2,       r=0 or 1

If r=0, then a= 2q, which is divisible by 2. So ,2q is even.

If r=1, then a= 2q+1, which is not divisible by 2.

Hence ,2q is odd.

Now,as ‘a’ is odd so it cannot be of the form 2q. This any odd positive integer 'a’ is of the form (2q+1).

Now, consider a²=(2q+1)²= 4q²+1+4q

[(x+y)=x²+y²+2xy]

a²= 4q²+4q+1

= 4(q²+q)+1  = 4m+1,      where m = q²+q

Hence, for some integer m,  the square of any odd integer is of the form 4m + 1.


HOPE THIS WILL HELP YOU....

Answer 2

$\huge \bf{ \red{hey}}$

$\huge{ \mathfrak{here \: is \: answer}}$

let a be any positive integer

then

b=8

0≤r<b

0≤r<4

r=0,1,2, 3

case 1.

r=0

a=bq+r

4q+0

(4q)^2

=> 16q^2

4(4q^2)

= let 2q^2 be m

4m

case 2.
r=1
a=bq+r

(4q+1)^2

(4q)^2+2*4q*1+(1)^2

16q^2+8q+1

4(4q^2+2q)+1.

let 4q^2+2q be. m

4m+1

case 3.

r=2

(4q+2)^2

(4q)^2+2*4q*2+(2)^2

16q^2+16q+4

4(4q^2+4q+1)

let 8q^2+4q+1 be m

4m

case4.

r=3
(4q+3)^2

(4q)^2+2*4q*3+(3)^2

16q^2+24q+9

16q^2+24q+8+1

4(4q^2+6q+1)+1

let 4q^2+6q+1 be m

4m+1

from above it is proved.

$\huge \boxed{ \boxed{ \pink{hope \: it \: helps}}}$

$\huge{ \green{thanks}}$