show/hide this revision's text 3 Simplified the proof

In the case $s=p/q$ with $|p| \geq 3$, one can prove that there are is no $3$-term AP in $P_s$, P_s$. The proof is by reducing to the case $s$ is an integer, 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 $\overline{A}=\overline{B}=\overline{C}=1$.L=K$.

Lemma 1. If $n^p$ is a $q$-th power in $K$, then $n$ is a $q$-th power in $\mathbf{Z}$.

We also remark that

Lemma 2. The integer $(A,B)=1$ (any common B$ is relatively prime factor of to $A$ and to $B$ has C$.

Proof. By symmetry, it suffices to divide prove $C$, by reasoning (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 integersof $L$), and similarly . By $(C,B)=1$. It (*)$ it follows that the subgroups $\langle \ell^{1/q} | C^{1/q}$. Thus $C/\ell \overline{A},\overline{C} in \rangle$ and $\langle mathbf{Q} \overline{B} cap \rangle$ intersect trivially in $K^{\times}/(K^{\times})^q$. In other words, the extensions $K(A^{1/q},C^{1/q})$ and overline{\mathbf{Z}} = \mathbf{Z}$ which contradicts $K(B^{1/q})$ are linearly disjoint(A,B,C)=1$. This proves Lemma 2.

If

By equation $(*)$, we had have $B^{1/q} K(B^{1/q}) \not\in K$, then there would exist subset K(A^{1/q},C^{1/q})$ which reads $\sigma \overline{B} \in \mathrm{Gal}(L/K)$ fixing langle \overline{A},\overline{C} \rangle$ in $A^{1/q},C^{1/q}$ but not K^{\times}/(K^{\times})^q$. We can thus write $B^{1/q}$. By applying B \equiv A^{\alpha} C^{\gamma} \pmod{(K^{\times})^q}$ for some $\sigma$ \alpha,\gamma \geq 0$. By a reasoning similar to the identity $A^{1/q}+C^{1/q}=2B^{1/q}$, Lemma 1, we get a contradiction. Thus $B^{1/q} \in K$ and by the lemma deduce that $B$ B/(A^{\alpha} C^{\gamma})$ is a $q$-th power in $\mathbf{Z}$, so that $A^{1/q}+C^{1/q} \\mathbf{Q}$ but since this fraction is in \mathbf{Z}$.

By a similar reasoning lowest terms (Lemma 2), we get that $A/C$ B$ is a $q$-th power in $\mathbf{Q}$, thus we can write $A=\lambda u^q$ and $C=\lambda v^q$ with \mathbf{Z}$.

Now let $\lambda,u,v \in \mathbf{Z}$. Replacing \sigma$ be an aribtrary element in $(*)$ we get \mathrm{Gal}(L/K)$. We have $\lambda^{1/q} \sigma(A^{1/q}) = \in zeta \mathbf{Q}$, so that cdot A^{1/q}$ and $\lambda$ is a \sigma(C^{1/q})=\zeta' \cdot C^{1/q}$ for some $q$-th power in roots of unity $\mathbf{Z}$ \zeta$ and $\zeta'$. Considering the real parts of both sides of $\sigma(*)$, we are donesee that necessarily $\zeta=\zeta'=1$. This shows that $L=K$ as requested.
show/hide this revision's text 2 edited body

In the case $s=p/q$ with $|p| \geq 3$, one can prove that there are no $3$-term AP in $P_s$, by reducing to the case $s$ is an integer, 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 $\overline{A}=\overline{B}=\overline{C}=1$.

Lemma. 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.

We also remark that $(A,B)=1$ (any common prime factor of $A$ and $B$ has to divide $C$, by reasoning in the ring of integers of $L$), and similarly $(C,B)=1$. It follows that the subgroups $\langle \overline{A},\overline{C} \rangle$ and $\langle \overline{B} \rangle$ intersect trivially in $K^{\times}/(K^{\times})^q$. In other words, the extensions $K(A^{1/q},C^{1/q})$ and $K(B^{1/q})$ are linearly disjoint.

If we had $B^{1/q} \not\in K$, then there would exist $\sigma \in \mathrm{Gal}(L/K)$ fixing $A^{1/q},C^{1/q}$ but not $B^{1/q}$. By applying $\sigma$ to the identity $A^{1/q}+C^{1/q}=2B^{1/q}$, we get a contradiction. Thus $B^{1/q} \in K$ and by the lemma $B$ is a $q$-th power in $\mathbf{Z}$, so that $A^{1/q}+C^{1/q} \in \mathbf{Z}$.

By a similar reasoning $A/C$ is a $q$-th power in $\mathbf{Q}$, thus we can write $A=\lambda u^q$ and $C=\lambda v^q$ with $\lambda,u,v \in \mathbf{Z}$. Replacing in $(*)$ we get $\lambda^{p/q} \lambda^{1/q} \in \mathbf{Q}$, so that $\lambda$ is a $q$-th power in $\mathbf{Z}$ and we are done.
show/hide this revision's text 1

In the case $s=p/q$ with $|p| \geq 3$, one can prove that there are no $3$-term AP in $P_s$, by reducing to the case $s$ is an integer, 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 $\overline{A}=\overline{B}=\overline{C}=1$.

Lemma. 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.

We also remark that $(A,B)=1$ (any common prime factor of $A$ and $B$ has to divide $C$, by reasoning in the ring of integers of $L$), and similarly $(C,B)=1$. It follows that the subgroups $\langle \overline{A},\overline{C} \rangle$ and $\langle \overline{B} \rangle$ intersect trivially in $K^{\times}/(K^{\times})^q$. In other words, the extensions $K(A^{1/q},C^{1/q})$ and $K(B^{1/q})$ are linearly disjoint.

If we had $B^{1/q} \not\in K$, then there would exist $\sigma \in \mathrm{Gal}(L/K)$ fixing $A^{1/q},C^{1/q}$ but not $B^{1/q}$. By applying $\sigma$ to the identity $A^{1/q}+C^{1/q}=2B^{1/q}$, we get a contradiction. Thus $B^{1/q} \in K$ and by the lemma $B$ is a $q$-th power in $\mathbf{Z}$, so that $A^{1/q}+C^{1/q} \in \mathbf{Z}$.

By a similar reasoning $A/C$ is a $q$-th power in $\mathbf{Q}$, thus we can write $A=\lambda u^q$ and $C=\lambda v^q$ with $\lambda,u,v \in \mathbf{Z}$. Replacing in $(*)$ we get $\lambda^{p/q} \in \mathbf{Q}$, so that $\lambda$ is a $q$-th power in $\mathbf{Z}$ and we are done.