In connection with this MO post (and without any applications / motivation whatsoever), here is an apparently difficult – but nice – problem.
For a non-zero real number $s$, consider the infinite sequence
$$ P_s := \{ 1^s, 2^s, 3^s, \ldots \} . $$
What is the number of terms, say $l(s)$, of the longest arithmetic
progression contained in this sequence?
For instance, since there exist three-term arithmetic progressions in
squares, but no such four-term progressions, we have $l(2)=3$.
It is easy to see that if $s$ is the reciprocal of a positive integer, then
$P_s$ contains an infinite arithmetic progression; hence we can write
something like $P(1/q)=\infty$. It can be shown that this is actually the
only case where $P_s$ contains an infinite progression. (This is certainly
non-trivial, but not that difficult either – in fact, in a different form
this was once posed as a problem on a Moscow State University math
competition).
If now $s$ is the reciprocal of a negative integer, then $P_s$ contains an
arithmetic progression of any preassigned length; this is a simple exercise.
Are there any other values of $s$ for which $l(s)$ is infinite?
Conjecture. For any real $s\ne 0$ which is not the reciprocal of an
integer, the quantity $l(s)$ is finite; that is, there exists an integer
$L>1$ (depending on $s$) such that $P_s$ does not contain $L$-term arithmetic
progressions.
Three-term progressions are not rare; say, for any integer $1<a<b<c$ with
$b>\sqrt{ac}$ there exists $s>0$ such that $\{a^s,b^s,c^s\}$ is an
arithmetic progression. However, I don't have any single example of a
four-term progression in a power sequence (save for the case where the
exponent is a reciprocal of an integer).
Is it true that $l(s)\le 3$ for any $s\ne 0$ which is not the reciprocal of an
integer?
Indeed, excepting the cases mentioned above and their immediate
modifications, I do not know of any $s$ such that $P_s$ contains two distinct
three-term progressions.
Is it true that if $s\ne p/q$ with integer $q\ge 1$ and
$p\in\{\pm1,\pm2\}$, then $P_s$ contains at most one three-term arithmetic
progression?
$$ $$
As a PS: I was once told that using relatively recent (post-Faltings) results
in algebraic number theory, one can determine $l(s)$ for $s$ rational. Can
anybody with the appropriate background confirm this?
Best Answer
In the case $s=p/q$ with $|p| \geq 3$, there is no $3$-term AP in $P_s$. The proof is by reducing to the case $s=p$, as follows.
Let $A=a^p$, $B=b^p$, $C=c^p$ such that $A^{1/q}+C^{1/q}=2B^{1/q} \quad (*) \quad$ and $(A,B,C)=1$. Let $K=\mathbf{Q}(\zeta_q)$ be the $q$-th cyclotomic field and $L=K(A^{1/q},B^{1/q},C^{1/q})$. Then $L/K$ is a finite abelian extension of exponent dividing $q$ and by Kummer theory, such extensions are in natural bijection with the finite subgroups of $K^{\times}/(K^{\times})^q$. The extension $L/K$ corresponds to the subgroup generated by the classes $\overline{A},\overline{B}, \overline{C}$ of $A,B,C$ in $K^{\times}/(K^{\times})^q$. In view of the following lemma, it suffices to prove $L=K$.
Lemma 1. If $n^p$ is a $q$-th power in $K$, then $n$ is a $q$-th power in $\mathbf{Z}$.
Proof. Assume $n^p=\alpha^q$ with $\alpha \in K$, then taking the norm we get $n^{p(q-1)}=N_{K/\mathbf{Q}}(\alpha)^q$. Since $\alpha$ is an algebraic integer, we get that $n^{p(q-1)}$ is a $q$-th power in $\mathbf{Z}$, and since $p(q-1)$ and $q$ are coprime, we get the result.
Lemma 2. The integer $B$ is relatively prime to $A$ and to $C$.
Proof. By symmetry, it suffices to prove $(A,B)=1$. Let $\ell$ be a prime number dividing $A$ and $B$. Then $\ell^{1/q}$ divides $A^{1/q}$ and $B^{1/q}$ in the ring $\overline{\mathbf{Z}}$ of all algebraic integers. By $(*)$ it follows that $\ell^{1/q} | C^{1/q}$. Thus $C/\ell \in \mathbf{Q} \cap \overline{\mathbf{Z}} = \mathbf{Z}$ which contradicts $(A,B,C)=1$. This proves Lemma 2.
By equation $(*)$, we have $K(B^{1/q}) \subset K(A^{1/q},C^{1/q})$ which reads $\overline{B} \in \langle \overline{A},\overline{C} \rangle$ in $K^{\times}/(K^{\times})^q$. We can thus write $B \equiv A^{\alpha} C^{\gamma} \pmod{(K^{\times})^q}$ for some $\alpha,\gamma \geq 0$. By a reasoning similar to Lemma 1, we deduce that $B/(A^{\alpha} C^{\gamma})$ is a $q$-th power in $\mathbf{Q}$ but since this fraction is in lowest terms (Lemma 2), we get that $B$ is a $q$-th power in $\mathbf{Z}$.
Now let $\sigma$ be an aribtrary element in $\mathrm{Gal}(L/K)$. We have $\sigma(A^{1/q}) = \zeta \cdot A^{1/q}$ and $\sigma(C^{1/q})=\zeta' \cdot C^{1/q}$ for some $q$-th roots of unity $\zeta$ and $\zeta'$. Considering the real parts of both sides of $\sigma(*)$, we see that necessarily $\zeta=\zeta'=1$. This shows that $L=K$ as requested.