Question

show that square of any positive integer is of the form 8q+1 for any integer q

Answer 1

By Euclid's division lemma, we have,
a=bq+r(where 0<=r<b)
so assuming the value of b as 8
a=8q+r(r=0, 1, 2, 3, 4, 5, 6, 7)
if r=0
a=8q
a^2=64q^2=8(8q^2)=8q
similarly it can be done for all values..

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

case 2.
r=1
a=bq+r

8q+1

case3.

r=2

a=bq+r

8q+2

case 4.

r=3

a=bq+r

8q+3

case 5.

r=4

a=bq+r

8q+4

case 6.

r=5

a=bq+r

8q+5

case7.

r=6

a=bq+r

8q+6

case 8

r=7

a=bq+r

8q+7

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

$\huge{ \green{thanks}}$