Question

Prove that for every positive integer n open can find n integers in arithmetical progression,all of them nontrivial powers of some integers, but one cannot find an infinite sequence with this property.

Answer 1

We prove the assertion by induction on $n$. For $n=3$, we can consider the numbers $1,25,49$.Suppose the assertion is true for some $n$ and let $a_i=b_i^{ki},i=1,2,...,n,$ be the terms of the progression having the common difference d. Let $b=a_n+d$ and let $k=1cm(k_1,k_2,...,k_n)$
Then the $n+1$ numbers $a_1{b}^k,a_2{b}^k,...,a_n{b}^k,b^{k+1}$ re in progression and all are power of some integers.Indeed, for all $i,1\le i\le n,$ there exists $d_i$ such that $k=k_i{d}_i$ and we obtain $a_i{b}^k=b_i^{ki}{b}^k=b_{bi}^{ki}{b}^{kidi}={\left(bi{b}^{di}\right)}^{ki}$ , as desired.For the second part,we need a result from Calculus:if $a,b>0$ then ${lim}_{n\to \infty }{\sum }_{k=1}^n\frac{1}{ak+b}=+\infty$ From this we deduce that if ${\left(a_n\right)}_{n\ge 1}$ is a progression of positive integers, then $\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{1}{ak}=+\infty$ Now suppose that ${\left(a_n\right)}_{n\ge 1}$ is a progression in which all terms are powers of some integers. Let S be the set of all positive integers greater than 1 that are powers of some integers. We will prove that $\sum _{aϵS}\frac{1}{a}\le \sum _{n\ge 2}\sum _{k\ge 2}\frac{1}{n^k}=\sum _{n\ge 2}\frac{1}{n^2}(1+\frac{1}{n}+\frac{1}{n^2}+\cdot \cdot \cdot )$
$=\sum _{n\ge 2}\frac{1}{n^2}\cdot \frac{1}{1-\frac{1}{n}}=\sum _{n\ge 2}\frac{1}{n(n-1)}=\sum _{n\ge 2}(\frac{1}{n-1}-\frac{1}{n})=1.$