Question

Prove that the square of any positive integer is of the form 5m or,5m+1 or,5m+4 for some integer m.

Answer 1

consider that m=2.square of 2=4, (2×2=4).
5m=5 (2)=10
4 not equals to 10

5m+1=5 (2)+1=11
4 not equals to 11

5m+4=5 (2)+4=14
14 is also not equals to 4.


If we take 5 as "m" then the square of 5=25,which satisfies the condition 5m
hence proved.

only 5 is the number which can satisfies the condition 5m. There is no other number which can satisfies the conditions 5m,5m+1,5m+4.

Answer 2

$\huge \bf{ \red{hey}}$

$\huge{ \mathfrak{ \blue{here \: is \: answer}}}$

let a be any positive integer

then

b= 5

a= bq+r

0≤r<b

0≤r<5

r= 0,1,2,3,4

case 1.

r=0

a= bq+r

5q+0

(5q)^2

25q^2

5(5q^2)

let 5q^2 be m

=> 5m

case 2.

r=1

a= 5q+1

(5q+1)^2

(5q)^2+2*5q*1+(1)^2

25q^2+10q+1

5(5q^2+2q)+1

let 5q^2+2q be m

=> 5m+1

case3.

r=2

a=5q+2
(5q+2)^2

(5q)^2+2*5q*2+(2)^2

25q^2+20q+4

5(5q^2+4q)+4

let 5q^2+4q be m

=> 5m+4

case4
r=

a=5q+3

(5q+3)^2

(5q)^2+2*5q*3+(3)^2

25q^2+30q+9

25q^2+30q+5+4

5(5q^2+6q+1)+4

let 5q^2+6q+1 be m

=>5m+4

case 5.

r=4

a=5q+4

(5q+4)^2

(5q)^2+2*5q*4+(4)^2

25q^2+40q+16

25q^2+40q+15+1

5(5q^2+8q+3)+1

let 5q^2+8q+3 be m

=> 5m+1

hence from above it is proved that square of any positive integer is of the form

5m, 5m+1 & 5m+4

HOPE\: IT \: HELPS

thanks