Prove that the square of an odd number is always one more than a multiple of 4

If we say n is any number, then we know 2n represents an even number - any number multiplied by 2 is always even. 2n+1 represents an odd number - adding 1 to an even number always gives an odd number (2n + 1)2 = (2n + 1)(2n + 1) = 4n2 + 2n + 2n + 1 = 4n2 + 4n + 1 = 4(n2 + n) + 1. Here 4(n2 + n) represents a multiple of four so we have a multiple of 4 plus 1. Hence the square of an odd number is always one more than a multiple of 4.

RR

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