Question

show that square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer q

Answer 1

hey

here is answer

let a be any positive integer

then

b=5

0≤r<b

0≤r<6

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

case 1.

r=0

a=bq+r

6q+0

(6q)^2

36q^2
6(6q^2)

let 6q^2 be m

=6m

case 2.
r=1
a=bq+r

(6q+1)^2

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

36q^2+12q+1

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

let 6q^2+2q be m

= 6m+1

case 3.

r=2

(6q+2)^2

36q^2+24q+4

6(6q^2+4q)+4

let 6q^2+4q be m

= 6m+4

case4.

r=3
(6q+3)^2

36q^2+36q+9

36q^2+36q+6+3
6(6q^2+6q+1)+3

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

= 6m+3

case 5.

r=4

(6q+4)^2

36q^2+48q+16

36q^2+48q+12+4

6(6q^2+8q+2)+4

let 6q^2+8q+4 be m

6m+4

case 6

r=5

(6q+5)^2

36q^2+60q+24+1

6(6q^2+10q+4)+1

let 6q^2+10q+4 bem

= 6m+1


note= i have taken m instead of q
from above it is proved.

hope it helps

thanks

Answer 2

Please check the attachment. Attachment is the solution of your question.................