Question

show that square of any odd integers is of the form 4m+1,For some integer m.

Answer 1

Step-by-step explanation:

Note :- I am taking q as some integer.

Let positive integer a be the any positive integer.

Then, b = 4 .

By division algorithm we know here

0 ≤ r < 4 , So r = 0, 1, 2, 3.

When r = 0

a = 4m

Squaring both side , we get

a² = ( 4m )²

a² = 4 ( 4m​²)

a² = 4q , where q = 4m²

When r = 1

a = 4m + 1

squaring both side , we get

a² = ( 4m + 1)²

a² = 16m² + 1 + 8m

a² = 4 ( 4m² + 2m ) + 1

a² = 4q + 1 , where q = 4m² + 2m

When r = 2

a = 4m + 2

Squaring both hand side , we get

a² = ​( 4m + 2 )²

a² = 16m² + 4 + 16m

a² = 4 ( 4m² + 4m + 1 )

a² = 4q , Where q = ​ 4m² + 4m + 1

When r = 3

a = 4m + 3

Squaring both hand side , we get

a² = ​( 4m + 3)²

a² = 16m² + 9 + 24m

a² = 16m² + 24m ​ + 8 + 1

a² = 4 ( 4m² + 6m + 2) + 1

a² = 4q + 1 , where q = 4m² + 6m + 2

Hence ,Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.

THANKS

#BeBrainly.

Answer 2

hey

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.


hope it helps

thanks