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N is an integer greater than 1
Prove algebraically that n²-2-(n-2)2 is always an even number.

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Answer:

Starting with the expression n² - 2 - (n - 2)², we can simplify it by expanding the second term using the formula for the square of a binomial:

n² - 2 - (n - 2)² = n² - 2 - (n² - 4n + 4)

Next, we can combine like terms:

n² - 2 - (n² - 4n + 4) = n² - n² + 4n - 6

Simplifying further, we get:

n² - 2 - (n - 2)² = 4n - 6

To prove that this expression is always even, we can show that it can be written in the form 2k, where k is an integer.

So, let's rewrite the expression as:

4n - 6 = 2(2n - 3)

Since 2n - 3 is an integer (because n is an integer greater than 1), we have shown that 4n - 6 can be written in the form 2k, where k is an integer.

Therefore, we have proved algebraically that n² - 2 - (n - 2)² is always an even number.

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