Question #143641
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.
1
Expert's answer
2020-11-11T12:25:37-0500

Let am=m2k,mZ+a_m=m^{2k}, m\in \Z^+ and an=n2k,nZ+a_n=n^{2k}, n\in \Z^+ be two numbers in this sequence.


n2km2k=(nkmk)(nk+mk)n^{2k}-m^{2k}=(n^k-m^k)(n^k+m^k)

If mm and nn are not consistent numbers the difference n2km2kn^{2k}-m^{2k} may be either odd or even.

The difference 42k22k4^{2k}-2^{2k} is even.

The difference 42k32k4^{2k}-3^{2k} is odd.


If mm and nn are not consistent numbers

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

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

{mis even nis odd =>{mkis even nkis 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}=>

=>{nkmkis odd nk+mkis odd =>(nkmk)(nk+mk)is odd=>\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 mm is odd, then n=m+1n=m+1 is even:

{mis odd nis even =>{mkis odd nkis 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}=>

=>{nkmkis odd nk+mkis odd =>(nkmk)(nk+mk)is odd=>\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.



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