Question

prove that if n odd integer then n^2+5 is even

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Neal

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An odd integer can be written in the form 2k + 1, where k is any integer.
So, if n is odd, we can write n = 2k + 1.
Then, n^2 + 5 = (2k + 1)^2 + 5 = 4k^2 + 4k + 1 + 5 = 4k^2 + 4k + 6.
We can factor out a 2 from each term to get 2(2k^2 + 2k + 3).
Since 2k^2 + 2k + 3 is an integer (because it’s a sum of integers), let’s denote it as m. So, we have n^2 + 5 = 2m.
A number is even if it can be written in the form 2m, where m is an integer. Therefore, n^2 + 5 is even.

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