Number theory: Is this argument correct and cube question First of all, is this argument correct?
Suppose $m < n$ are integers. Then for every $k \in \mathbb{N}$ 
$$m + k < n + k.$$
What I did: Suppose that $m + k \ge n + k$, the by the cancelation law $m\ge n$. What gives us a contradiction.
Is this right?
Another problem is, I have to show that $2^n + 1$ is not a cube for every $n\ge 0$. I think that I have to use induction, but I don't know how.
Thanks!
 A: The first problem reads as one of those "obvious" proofs, but it is not a result that should be treated trivially. I imagine you should try to prove it more rigorously. Cite some of the basic axioms you have available to you in number theory, or use some properties that you know about inequalities. I wouldn't expect that the cancelation law would be something available to you in introductory number theory, but if it is then I buy your proof. 
Induction is certainly a way to prove your second result. With induction you first prove the base case. For you this means show that $2^0+1$ is not a cube. That should be pretty easy. Then suppose $2^n+1$ is not a cube for all $n \in \{0,1,\ldots, k\}$. You now want to know what can be said of the quantity $2^{k+1}+1$. Try rewriting the quantity to coax out something useful. For example, rewrite $2^{k+1}+1$ in a way so that you can see $2^k+1$. You already know $2^k+1$ is not a cube, namely that it cannot be written as some integer $m^3.$ $\mathbf{Peter}$ outlined a great way to prove this in the comment section.
However I see a way to do that proof through infinite descent which I've never gotten to do before! It is a terrible way to prove this result but here's how it would look. Suppose for some $k>0$ there exists an $m$ such that $2^k+1 = m^3$. Clearly the LHS is odd so $m$ must too be odd. Suppose $m = 2n_1+1$, expand the RHS and simplify. You'll get $$2^k = 8n_1^3+12n_1^2+6n_1$$ or $$2^{k-1} = 2(2n_1^3+3n_1^2)+3n_1$$ if $k = 1$ the LHS would be $1$ and no $n_1 \geq 0 $ can satisfy the equation. You can verify this by trying $n_1=0,n_1=1$ and deducing the equality fails for all $n_1>0$. So $k$ must be greater than $1$, meaning $2^{k-1}$ is even. This in turn means $2(2n_1^3+3n_1^2)+3n_1$ is even, so $n_1$ cannot be odd. Now we know $n_1 = 2n_2$ for some other integer $n_2$. Plugging in $2n_2$ in place of $n_1$ yields $$2^{k-1} = 2(16n_2^3+12n_2^2)+6n_2$$ or $$2^{k-2} = 2(8n_2^3+6n_2^2)+3n_2$$ Again we see if $k = 2 $ we get a contradiction. So we have to have $n_2$ be even as well and some $n_3$ such that $n_2 = 2n_3$. This process can always be repeated $k$ times to get $$1 = 2(a_kn_k^3+b_kn_k^2)+3n_k$$ where $a_k, b_k$ are just the constants that appear on the $k$-th iteration. In this way we always end up with an integer $n_k$ that cannot exist, meaning $m$ cannot exist. So we conclude for all $k \in \mathbb{N}$ that $2^k+1 \neq m^3$ for any $m \in \mathbb{N}$.
