Question

Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4

Answer 1

Let the integer be x

The square of its integer is x² 

Let x be an even integer

x = 2q + 0

x² = 4q²

When x is an odd integer

x = 2k + 1

x² = (2k + 1)²

= 4k² + 4k + 1

= 4k (k + 1) + 1

= 4q + 1 ......[q = k(k + 1)]

It is divisible by 4

Hence it is proved