## Discrete Math Problem

Arrangements, sorting, packing, partitions, critical path analysis, networks, graphs, ...
Jkepler
Posts: 5
Joined: Mon Jan 31, 2011 7:33 pm

### Discrete Math Problem

Prove the following about integers: x is an odd integer iff x2 + 2x + 1 is an even integer.
"There is no dark side in the moon, really. Matter of fact, it's all dark."

hk
Posts: 10110
Joined: Sun Mar 26, 2006 9:34 am
Location: Haren, Netherlands

### Re: Discrete Math Problem

Do you mean x2+2x+1?
If so I suspect that you copied the question from somewhere else.
Anyhow: try to rewrite x2+2x+1 (with brackets) and it will become pretty obvious IMHO.

Jkepler
Posts: 5
Joined: Mon Jan 31, 2011 7:33 pm

### Re: Discrete Math Problem

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 x2 + 2x + 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!
"There is no dark side in the moon, really. Matter of fact, it's all dark."

thundre
Posts: 356
Joined: Sun Mar 27, 2011 9:01 am

### Re: Discrete Math Problem

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

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 x2 + 2x + 1 is even, the other starting with the opposite.

Ellenion
Posts: 16
Joined: Fri Jul 14, 2017 10:09 pm

### Re: Discrete Math Problem

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.