*x*is an odd integer iff

*x*2 + 2

*x*+ 1 is an even integer.

Prove the following about integers: *x* is an odd integer iff *x*2 + 2*x* + 1 is an even integer.

Do you mean x^{2}+2x+1?

If so I suspect that you copied the question from somewhere else.

Anyhow: try to rewrite x^{2}+2x+1 (with brackets) and it will become pretty obvious IMHO.

If so I suspect that you copied the question from somewhere else.

Anyhow: try to rewrite x

Yes, correct on both counts. I meant that to be an exponent, and I *had* copied and pasted it from elsewhere.

Anyway, here's my solution:

Factoring tells us that*x*^{2} + 2*x* + 1 = (*x* + 1)^{2}

So,*x* is odd iff *x* + 1 is even iff (*x* + 1)^{2} is even.

Would this be an acceptable proof?

Thanks again!

Anyway, here's my solution:

Factoring tells us that

So,

Would this be an acceptable proof?

Thanks again!

It depends how much rigor is expected in the class.Jkepler wrote:Would this be an acceptable proof?

To prove "A iff B", you need to prove "B implies A" and "not B implies not A". You could do it in 2 separate sections, one starting with the statement

Proof:

It can be seen that $x^2+2x+1$ is even, as given in the problem. Therefore, $x^2+2x$ is odd, since an even number minus $1$ is odd. Then, we know that any integer multiplied by $2$ will be even, and an odd number minus an even number will be odd. Therefore, $x^2$ is odd. Since squares of odd numbers are the only odd squares, then $x$ has to be odd.

It can be seen that $x^2+2x+1$ is even, as given in the problem. Therefore, $x^2+2x$ is odd, since an even number minus $1$ is odd. Then, we know that any integer multiplied by $2$ will be even, and an odd number minus an even number will be odd. Therefore, $x^2$ is odd. Since squares of odd numbers are the only odd squares, then $x$ has to be odd.