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