Show that the square of any positive integer is either of the form 4m or 4m+1 for some integer n
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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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