Question

Show that the square of any odd positive integer is in the form of 8m+1

Answer 1

According to Euclid division lemma , a = bq + r where 0 ≤ r < b
Here b = 8 possible remainder = 1, 2, 3,4,5,6 and 7
Then, a = 8q + r

Case 1
when r = 1 , a = 8q + 1
squaring on both sides,
a² = (8q + 1)² = 64²q² + 16q + 1 = 8(8q² + 2q) + 1
= 8m + 1 , where m = 8q² + 2q

case 2
when r = 2 , a = 8q + 2
squaring on both sides,
a² = (8q + 2)² = 64q² + 32q + 4 ≠ 8m +1

Case 3
when r = 3 , a = 8q + 3
squaring on both sides,
a²= (8q + 3)²= 64q²+48q + 9 =64q²+48q +8+1=8(8q² + 6q + 1) + 1
= 8m + 1 , where m = 8q² + 6q + 1

You can see that at every odd values of r square of a is in the form of 8m +1
But at every even Values of r square of a isn't in the form of 8m +1 .
Also we know, a = 8q +1 , 8q +3 , 8q + 5 , 8q +7 are not divisible by 2 means these all numbers are odd numbers
Hence , it is clear that square of an odd positive is in form of 8m +1

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<8

r=0,1,2, 3,4,5,6,7

case 1.

r=0

a=bq+r

8q+0

(8q)^2

=> 64q^2

8(8q^2)

= let 2q^2 be m

8m

case 2.
r=1
a=bq+r

(8q+1)^2

(8q)^2+2*8q*1+(1)^2

64q^2+16q+1

8(8q^2+2q)+1.

let 8q^2+2q be. m

8m+1

case 3.

r=2

(8q+2)^2

(8q)^2+2*8q*2+(2)^2

64q^2+32q+4

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

let 8q^2+4q be m

8m+4

case4.

r=3
(8q+3)^2

(8q)^2+2*8q*3+(3)^2

64q^2+48q+9

64q^2+48q+8+1

8(8q^2+6q+1)+1

let 8q^2+6q+1 be m

8m+1

case 5.

r=4

(8q+4)^2

64q^2+64q+16
8(8q^2+8q+2)

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

8m

case 6

r=5

(8q+5)^2

64q^2+80q+25

64q^2+80q+24+1

8(8q^2+10q+3)+1

let 8q^2+10q+3 be m

8m+1

case 7

r=6

(8q+6)^2

64q^2+96q+36

64q^2+96q+32+4

8(8q^2+12q+4)+4

let 8q^2+12q+4 be m..

8m+4

case8.

r=7

(8q+7)^2

64q^2+112q+49

64q^2+112q+48+1

8(8q^2+14q+6)+1

let 8q^2+14q+6 be m

8m+1

from above it is proved.

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

$\huge{ \green{thanks}}$