Answer to Question #143641 in Combinatorics | Number Theory for 1805040444b

Question

Consider the following sequence of successive numbers of the 2k-th power:
1, 2^2k, 3^2k, 4^2k, 5^2k, ...
Show that the difference between the numbers in this sequence is odd for all k ∈ N.

Answer 1

Let $a_m=m^{2k}, m\in \Z^+$ and $a_n=n^{2k}, n\in \Z^+$ be two numbers in this sequence.

$$n^{2k}-m^{2k}=(n^k-m^k)(n^k+m^k)$$

If $m$ and $n$ are not consistent numbers the difference $n^{2k}-m^{2k}$ may be either odd or even.

The difference $4^{2k}-2^{2k}$ is even.

The difference $4^{2k}-3^{2k}$ is odd.

If $m$ and $n$ are not consistent numbers

Suppose $n=m+1, m\in \Z^+$

If $m$ is even, then $n=m+1$ is odd:

$$\begin{cases} m &\text{is even } \\ n &\text{is odd } \end{cases}=>\begin{cases} m^k &\text{is even } \\ n^k &\text{is odd } \end{cases}=>$$

$$=>\begin{cases} n^k-m^k &\text{is odd } \\ n^k+m^k &\text{is odd } \end{cases}=>(n^k-m^k)(n^k+m^k) \text{is odd}$$

If $m$ is odd, then $n=m+1$ is even:

$$\begin{cases} m &\text{is odd } \\ n &\text{is even } \end{cases}=>\begin{cases} m^k &\text{is odd } \\ n^k &\text{is even } \end{cases}=>$$

$$=>\begin{cases} n^k-m^k &\text{is odd } \\ n^k+m^k &\text{is odd } \end{cases}=>(n^k-m^k)(n^k+m^k) \text{is odd}$$

Therefore  the difference between the consistent numbers in this sequence is odd for all k ∈ N.