Question

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?

Answer 1

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

Explanation:

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

`n^2+n+1` is odd

Explanation:

• Even numbers are numbers of the form `2k` for some integer `k`.

• Odd numbers are numbers of the form `2k+1` for some integer `k`.

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

Is odd.

Explanation:

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

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