Question

Prove that the square of any positive integer of the the form 5m+1 will leave a remainder 1. When divided by 5 for some integer m

Answer 1

Since; 
(5m+1)2 = 25m2+1+10m

Here the term 25m2 is divisible by 5 and the term 10m is also exactly divisible by 5.
Therefore it leaves the remainder 1.

Putting m = 1, we get; (5×1+1)2 = 62 = 36
And 36 when divided by 5, leaves the remainder 1.

Therefore square of any positive integer is of the form 5m+1 will leave the remainder 1 when divided by 5 for some integer m.

plzz mark as brainliest answer !!!

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

from above it is proved.

hope it helps

thanks