I think this follows (with a very weak bound) from van der Waerden's theorem on gives at least an arithmetic progressionsprogression (not consecutive elements) consisting of $n$-th powers. Let $k,n$ be given and $p$ a sufficiently large prime. If $g$ is a primitive root mod $p$, then consider the following coloring of the reduced residue system: $x\mapsto c$ if $x\equiv g^{ny+c}\pmod{p}$ for some $y$ and $0\leq c \lt n$ . This is a coloring of $0,1,\dots,p-1$ with $n$ colors. By van der Waerden's theorem, there is an AP of length $k$, i.e., $y,y+H,\dots,y+(k-1)H$ get the same color, $c$. If we divide by $g^c$, we get an AP $z,z+h,\dots,z+(k-1)h$ of length $k$, consisting $n$-th powers. Finally, we divide by $h$ and get $k$ consecutive $n$-th powers.
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I think this follows (with a very weak bound) from van der Waerden's theorem on arithmetic progressions. Let $k,n$ be given and $p$ a sufficiently large prime. If $g$ is a primitive root mod $p$, then consider the following coloring of the reduced residue system: $x\mapsto c$ if $x\equiv g^{ny+c}\pmod{p}$ for some $y$ and $0\leq c \lt n$ . This is a coloring of $0,1,\dots,p-1$ with $n$ colors. By van der Waerden's theorem, there is an AP of length $k$, i.e., $y,y+H,\dots,y+(k-1)H$ get the same color, $c$. If we divide by $g^c$, we get an AP $z,z+h,\dots,z+(k-1)h$ of length $k$, consisting $n$-th powers. Finally, we divide by $h$ and get $k$ consecutive $n$-th powers. |
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I think this follows (with a very weak bound) from van der Waerden's theorem on arithmetic progressions. Let $k,n$ be given and $p$ a sufficiently large prime. If $g$ is a primitive root mod $p$, then consider the following coloring of the reduced residue system: $x\mapsto c$ if $x\equiv g^{ny+c}\pmod{p}$ for some $y$ and $0\leq c |
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