Question

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

Answer 1

Let b be 2 ...
Then by EDL ,
a=2q+r ...
There will be two cases when remainder is 1 and 0
But it is saying odd integer ... So only
A=2q+1 will be taken ....
Now squaring both side ...
A^2= 4q^2+4q+1 ....
Now a= 4m+1. (Where m = q^2+q)

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.

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

$\huge{ \green{thanks}}$