Question

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

Answer 1

Answer:

We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.

Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5

Thus we have:

(6 m +1)2 = 36 m2 + 12 m + 1 = 6 (6 m2 + 2 m) + 1 = 6 q + 1,

q is an integer

(6 m + 3)2 = 36 m2 + 36 m + 9 = 6 (6 m2 + 6 m + 1) + 3 = 6 q + 3,

q is an integer

(6 m + 5)2 = 36 m2 + 60 m + 25 = 6 (6 m2 + 10 m + 4) + 1 = 6 q + 1,

q is an integer.

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Answer 2

$\underline{\underline{\huge{\pink{\tt{\textbf Answer :-}}}}}$

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