Say whether the following is true or false and support your answer by a proof: For any integer n, the number n2+n+1 is odd?

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Answer 1 · 2017-08-25T15:28:25

$n^2+n+1$ is always odd

If $n$ is odd:
$n^2$ is also odd
$n^2+n$ (sum of two odds) is even
$n^2+n+1$ is odd

If $n$ is even:
$n^2$ is also even
$n^2+n$ (sum of two evens) is even
$n^2+n+1$ is odd

Answer 2 · 2017-08-25T19:43:06

$n^2+n+1$ is odd

Every integer is either odd or even.

$color(white)()$
Case $n$ even

If $n$ is even, then there is some integer $k$ such that $n = 2k$

Then:

$n^2+n+1 = (2k)^2+2k+1$

$color(white)(n^2+n+1) = 4k^2+2k+1$

$color(white)(n^2+n+1) = 2(2k^2+k)+1$

$color(white)(n^2+n+1) = 2k_1+1$

where $k_1 = 2k^2+k$ is an integer.

So $n^2+n+1$ is odd.

$color(white)()$
Case $n$ odd

If $n$ is odd, then there is some integer $k$ such that $n = 2k+1$

Then:

$n^2+n+1 = (2k+1)^2+(2k+1)+1$

$color(white)(n^2+n+1) = 4k^2+2k+1+2k+1+1$

$color(white)(n^2+n+1) = 2(2k^2+2k+1)+1$

$color(white)(n^2+n+1) = 2k_2+1$

where $k_2 = 2k^2+2k+1$ is an integer.

So $n^2+n+1$ is odd.

$color(white)()$
Conclusion

So regardless of whether $n$ is even or odd, $n^2+n+1$ is odd.

Answer 3 · 2017-08-25T20:48:27

Is odd.

$n^2+n+1 = n(n+1)+1$

Note that $n(n+1)$ is always even so $n(n+1)+1$ is odd