Question

prove square of any positive integer is in the form 5q 5q+1 5q+4 for dome integer q

Answer 1

Let x be any integer
Then x = 5m or x = 5m+1 or x = 5m+4 for integer x.
If x = 5m, x2 = (5m)2 = 25m2 = 5(5m2) = 5n (where n = 5m2 )
If x = 5m+1, x2 = (5m+1)2 = 25m2+10m+1 = 5(5m2+2m)+1 = 5n+1 (where n = 5m2+2m )
If x = 5m+4, x2 = (5m+4)2 = 25m2+40m+16 = 5(5m2+8m+3)+1 = 5n+1 (where n = 5m2+8m+3 )
∴in each of three cases x2 iseither of the form 5n or 5n+1 for integer n.

Answer 2

hey

here is answer

let a be any positive integer

then

b=5

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=bq+r

(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

case 3.

r=2

(5q+2)^2

25q^2+20q+4

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

let 5q^2+4q be m

= 5m+4

case4.

r=3
(5q+3)^2

25q^2+30q+9

25q^2+30q+5+4

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

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

= 5m+4

case 5.

r=4

(5q+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


note= i have taken m instead of q

from above it is proved.

hope it helps

thanks