Question

show that square of any positive integer is of the form 4m or 4m +2, where m is any integer

Answer 1

set A be any positive integer which is divided by 4 leaving quotient and r as remainder

a=bq+r,0 < or equal to r,r <b
for r=o
a=(4m+0)^2
=16m
=4(4m)
=4M,where M = 4m


for r=1
a=(4m+1)^2
=(4m)^2+1^2+2 (4m)
=16m^2+1+8m
=4 (4m^2+2)+1
=4M+1,where M=4m^2+2

for r=2
=(4m+2)^2
=(4m)^2 + 2^2 + 2(4m)(2)
=16m^2 + 4 + 16m
=4 (4m +1+4m)
=4M, where M=4m + 1 + 4m

therefore any positive integer is of the form 4m or 4m +2, where m is any integer

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.

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

$\huge{ \green{thanks}}$