Question

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

Answer 1

let X be any positive integer in the form 4m, 4m+ 1
we know that
a = bq+ r
put b =4
acc. to question
a^2 = (4q + 0) ^2
= (16q^2)
= 4. 4q^2
=4m
where m = 4q^2
2) a^2 = (4q +1)^2
= 16q^2 +1 + 4q
= 4 (4q^2 + q) +1
= 4m + 1
where m = 4q^2 + q
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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<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.

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

$\huge{ \green{thanks}}$