Question

Show that square of any positive odd integer is of the form 8m+1 for some integer m???

Answer 1

let AB be any positive odd interger and q and r be their unique exist
so a is equal to bq pluse r.
let r is equal to 1,2,3,4.
so on the place of 8m let be 4m .
caseI let r be equal to 1
aequal to 4m multiplyby 1 pluse remaimder .
so a is equal to 4m taking squalre on both side you can do like this.

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<8

r=0,1,2, 3,4,5,6,7

case 1.

r=0

a=bq+r

8q+0

(8q)^2

=> 64q^2

8(8q^2)

= let 2q^2 be m

8m

case 2.
r=1
a=bq+r

(8q+1)^2

(8q)^2+2*8q*1+(1)^2

64q^2+16q+1

8(8q^2+2q)+1.

let 8q^2+2q be. m

8m+1

case 3.

r=2

(8q+2)^2

(8q)^2+2*8q*2+(2)^2

64q^2+32q+4

8(8q^2+4q)+4

let 8q^2+4q be m

8m+4

case4.

r=3
(8q+3)^2

(8q)^2+2*8q*3+(3)^2

64q^2+48q+9

64q^2+48q+8+1

8(8q^2+6q+1)+1

let 8q^2+6q+1 be m

8m+1

case 5.

r=4

(8q+4)^2

64q^2+64q+16
8(8q^2+8q+2)

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

8m

case 6

r=5

(8q+5)^2

64q^2+80q+25

64q^2+80q+24+1

8(8q^2+10q+3)+1

let 8q^2+10q+3 be m

8m+1

case 7

r=6

(8q+6)^2

64q^2+96q+36

64q^2+96q+32+4

8(8q^2+12q+4)+4

let 8q^2+12q+4 be m..

8m+4

case8.

r=7

(8q+7)^2

64q^2+112q+49

64q^2+112q+48+1

8(8q^2+14q+6)+1

let 8q^2+14q+6 be m

8m+1

from above it is proved.

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

$\huge{ \green{thanks}}$