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

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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.

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