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Given that n is an interger and that n>1 prove algebraically that n^2-(n-2)^2-2 is always an even number

User JMoravitz
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Answer and Step-by-step explanation:

If n is assumed to be an even number, then obviously the expression will be even because the square of an even number is even and subtracting 2 from an even number makes it remain an even number. Therefore, let's assume n is not an even number, then n = 2x + 1 where x > 0. We have:

(2x + 1)² – (2x + 1 – 2)² – 2 = (2x + 1)² – (2x – 1)² – 2 = 4x² + 4x + 1 – (4x² – 4x + 1) – 2 = 4x² + 4x + 1 – 4x² + 4x – 1 – 2 = 8x – 2 = 2(4x – 1). Since the product is a multiple of 2, then the expression is even if n is odd. Therefore the expression is always even.

User Yoones
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