Question

show that the square of any positive integer is of the form 5k, 5k+1 and 5k+4 for some integer k​

Answer 1

Let x be any positive integer Then x = 5k or x = 5k+1 or x = 5k+4 for integer x. If x = 5k, x2 = (5k)2 = 25k2 = 5(5k2) = 5n (where n = 5k2 ) If x = 5k+1, x2 = (5k+1)2 = 25k2+10k+1 = 5(5k2+2k)+1 = 5n+1 (where n = 5k2+2k ) If x = 5k+4, x2 = (5k+4)2 = 25q2+40q+16 = 5(5k2 + 8k+ 3)+ 1 = 5n+1 (where n = 5k2+8k+3 ) ∴in each of three cases x2 is either of the form 5k or 5k+1 or 5k+4 and for integer k.

Answer 2

Let positive integer a = 5m+ r , By division algorithm we know here 0 ≤ r < 5 , So

When r = 0

a = 5m  

Squaring both side , we get

a2 = ( 5m)2

a2 = 5 ( 5m​2)

a2 = 5q, where q = 5m2

When r = 1

a = 5m + 1

squaring both side , we get

a2 = ( 5m + 1)2

a2 = 25m2 + 1 + 10m  

a2 = 5 ( 5m2 + 2m) + 1  

a2 = 5q + 1 , where q = 5m2 + 2m

When r = 2

a = 5m + 2  

Squaring both hand side , we get

a2 = ​( 5m + 2)2

a2 = 25m2 + 5 + 20m  

a2 = 5 ( 5m2 + 4m + 5)

a2 = 5q , Where q = ​ 5m2 + 5m + 1

When r = 3  

a = 5m + 3

Squaring both hand side , we get

a2 = ​( 5m + 3)2

a2 = 25m2 + 9 + 30m  

a2 = 25m2 + 30m ​ + 10- 1

a2 = 5 ( 5m2 + 6m + 2) - 1

a2 = 5q -1 , where q = 5m2 + 6m + 2  

Hence  

Square of any positive integer is in form of 5q or 5q + 4. , where q is any integer.