Question

prove that the square of any position integer is of the from 5q, 5q+1,5q+4 for some integer q.​

Answer 1

Let positive integer a = 5m+ r.

By division algorithm we know here 0 ≤ r < 5.

                                                           

When r = 0

a = 5m  

Squaring both sides , we get

a² = ( 5m)²

a² = 5 ( 5m​²)

a² = 5q, where q = 5m²

                                                           

When r = 1

a = 5m + 1

Squaring both sides , we get

a² = ( 5m + 1)2²

a² = 25m² + 1 + 10m  

a² = 5 ( 5m² + 2m) + 1  

a²= 5q + 1 , where q = 5m² + 2m

                                                             

When r = 2

a = 5m + 2  

Squaring both sides , 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 sides, we get

a² = ​( 5m + 3)2²

a² = 25m² + 9 + 30m  

a² = 25m² + 30m ​ + 10- 1

a² = 5 ( 5m² + 6m + 2) - 1

a² = 5q -1 , where q = 5m² + 6m + 2  

${\boxed{\bold{Therefore \ Square \ of \ any \ integer \ is \ of \ the \ form \ 5q, \ 5q+1 \ 5q+4 \ for \ some \ integer \ q.}}}$