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?

  • $\begingroup$ Regarding $P_s$ containing at most one $3$-term AP, I guess you want to assume $a,b,c$ are coprime (otherwise you could take any AP and multiply it with $k^s$). $\endgroup$ – François Brunault Oct 24 '11 at 14:17
  • $\begingroup$ @Francois: absolutely. $\endgroup$ – Seva Oct 24 '11 at 14:31
  • $\begingroup$ This question is similar to this one: mathoverflow.net/questions/59471/… $\endgroup$ – Kevin O'Bryant Oct 24 '11 at 17:14

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.

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    $\begingroup$ Thanks! We thus have $$ l(p/q) = \begin{cases} \infty &\text{if}\ |p|=1, \\ 3 &\text{if}\ |p|=2, \\ 2 &\text{if}\ |p|\ge 3. \end{cases} $$ $\endgroup$ – Seva Oct 25 '11 at 8:35

When s is an integer, you are asking about three-term arithmetic progressions among nth powers. There are none when n > 2! This was a 1952 conjecture of Denes and is now a theorem of Darmon and Merel, part of the wave of Diophantine results that followed in the wake of Wiles's work on modularity. (Indeed, the existence of a 3-term AP among nth powers is tantamount to a solution to a^n - 2 b^n = c^n, which is manifestly in the same ballpark as Fermat's equation.)

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  • $\begingroup$ Thanks for the reply - but, frankly, I was aware of that paper of Darmon and Merel. Perhaps, I should have mentioned it in my original post, but somehow I feel it is already somewhat longish... What I do not know is whether Darmon-Merel extends onto the rational case. (I was once told it does, but would be happy if someone could confirm this.) Maybe, on this occasion: another fact I have not mentioned is that $l(-s)=l(s)$; and so one can confine to the case $s>0$. $\endgroup$ – Seva Oct 24 '11 at 15:15
  • $\begingroup$ @Seva : I think that one can reduce the rational case to the integral case using some Galois theory of Kummer extensions. The idea is that if $a^{p/q}+c^{p/q}=2 b^{p/q}$ then we get other identities by applying elements of the Galois group. From this it should be possible to prove that $a,b,c$ are $q$-th powers. Well, this is very rough, so one should check the details. $\endgroup$ – François Brunault Oct 24 '11 at 15:42

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