Question

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

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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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.

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