So I started with if $n^3$ is odd, then $n^2$ is odd and I used a direct proof. $n^3 = n^2 \cdot n$ and the product of 2 odd integers is odd so $n^2$ is odd. Is this an okay proof?

Then I have to prove that if $n^2$ is odd then $1-n$ is even, which I did.

Then I need to prove that if $1-n$ is even, then $n^2+1$ is even. Here I get stuck. With direct proof I assume that $1-n$ is even, so $1-n =2k$. Can I do the following: $$n=-2k+1$$ $$n^2+1= (-2k+1)^2+1 \quad \implies \quad n^2k+1=2(2k^2-2k+1)$$ which is even?

And finally following a similar logic, if $n^2+1$ is even then $n^3$ is odd. $$n^2+1=2k \quad \implies \quad \begin{cases} n^2=2k-1 \\ n^3=n^2\cdot n \end{cases}$$ And since the product of $2$ odd numbers are odd numbers, I can say that $n=2k+1$, so then I replace everything and say that $n^2 \cdot n=(2k-1)(2k+1)=2(2k)-1$ which is odd?

  • 2
    $\begingroup$ I think the last two "odd"s in the title should be "even", right? $\endgroup$ – rogerl Oct 21 '16 at 16:17
  • $\begingroup$ In the first part, you say the product of two odd integers is odd. Have you implicitly used the fact that $n$ must be odd? If so, you should probably make that explicit and explain why it is so. That issue comes up in the fourth part as well. $\endgroup$ – rogerl Oct 21 '16 at 16:18
  • $\begingroup$ If $n^2$ is odd, it will be awfully hard for $n^2+1$ to be odd... $\endgroup$ – TravisJ Oct 21 '16 at 16:20
  • $\begingroup$ oh sorry yes the last 2 should be even $\endgroup$ – Yanni Papaioannou Oct 21 '16 at 16:26
  • $\begingroup$ @YanniPapaioannou - As you can see, your post is much more readable with 'mathjax' added. Next time you ask a question, please have a go at learning to use it. It's an essential tool for writing good answers and questions. $\endgroup$ – Myridium Oct 21 '16 at 16:28

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