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Only a partial answer for the special case of an integer $n$ with Galois radius $r$ of type $(a,b)$ where $\min(a,b)=1$.

Denote by $\mathbb{P}_{k}$ the set of $k$-th powers of primes and suppose $b>a$, so that $(n-r,n+r)=(p,q^{b})$ with $p$ and $q$ odd primes. Provided $n>5^{b}$, consider the closest odd element $q'$ of $\mathbb{P}_{b}$ to $p$ less than $p$ and set $2t=p-q'$. Now consider the closest prime $p'$ to $n+r+2t$ greater than $n+r+2t$. Then $2r':=p'-q'$ is twice a Galois radius of $n$ of type $(b,a)$ and we may set $s:=\vert r-r'\vert$, which that way fulfills the required conditions to be an inversion shift of $r$. The same line of reasoning applies mutatis mutandis for the case $b<a$.

Edit: actually, there seems to be a problem with this reasoning, since $r'$ is a Galois radius of type $(b,a)$ of an integer $n'$ which may differ from $n$. Hence one must prove $n$ is large enough to ensure $n=n'$.

Only a partial answer for the special case of an integer $n$ with Galois radius $r$ of type $(a,b)$ where $\min(a,b)=1$.

Denote by $\mathbb{P}_{k}$ the set of $k$-th powers of primes and suppose $b>a$, so that $(n-r,n+r)=(p,q^{b})$ with $p$ and $q$ odd primes. Provided $n>5^{b}$, consider the closest odd element $q'$ of $\mathbb{P}_{b}$ to $p$ less than $p$ and set $2t=p-q'$. Now consider the closest prime $p'$ to $n+r+2t$ greater than $n+r+2t$. Then $2r':=p'-q'$ is twice a Galois radius of $n$ of type $(b,a)$ and we may set $s:=\vert r-r'\vert$, which that way fulfills the required conditions to be an inversion shift of $r$. The same line of reasoning applies mutatis mutandis for the case $b<a$.

Edit: actually, there seems to be a problem with this reasoning, since $r'$ is a Galois radius of type $(b,a)$ of an integer $n'$ which may differ from $n$. Hence one must prove $n$ is large enough to ensure $n=n'$. Maybe this issue can be resolved considering an integer $M$ such that $r'\equiv -r\pmod M$.

Only a partial answer for the special case of an integer $n$ with Galois radius $r$ of type $(a,b)$ where $\min(a,b)=1$.

Denote by $\mathbb{P}_{k}$ the set of $k$-th powers of primes and suppose $b>a$, so that $(n-r,n+r)=(p,q^{b})$ with $p$ and $q$ odd primes. Provided $n>5^{b}$, consider the closest odd element $q'$ of $\mathbb{P}_{b}$ to $p$ less than $p$ and set $2t=p-q'$. Now consider the closest prime $p'$ to $n+r+2t$ greater than $n+r+2t$. Then $2r':=p'-q'$ is twice a Galois radius of $n$ of type $(b,a)$ and we may set $s:=\vert r-r'\vert$, which that way fulfills the required conditions to be an inversion shift of $r$. The same line of reasoning applies mutatis mutandis for the case $b<a$.

Only a partial answer for the special case of an integer $n$ with Galois radius $r$ of type $(a,b)$ where $\min(a,b)=1$.

Denote by $\mathbb{P}_{k}$ the set of $k$-th powers of primes and suppose $b>a$, so that $(n-r,n+r)=(p,q^{b})$ with $p$ and $q$ odd primes. Provided $n>5^{b}$, consider the closest odd element $q'$ of $\mathbb{P}_{b}$ to $p$ less than $p$ and set $2t=p-q'$. Now consider the closest prime $p'$ to $n+r+2t$ greater than $n+r+2t$. Then $2r':=p'-q'$ is twice a Galois radius of $n$ of type $(b,a)$ and we may set $s:=\vert r-r'\vert$, which that way fulfills the required conditions to be an inversion shift of $r$. The same line of reasoning applies mutatis mutandis for the case $b<a$.

Edit: actually, there seems to be a problem with this reasoning, since $r'$ is a Galois radius of type $(b,a)$ of an integer $n'$ which may differ from $n$. Hence one must prove $n$ is large enough to ensure $n=n'$.

Only a partial answer for the special case of an integer $n$ with Galois radius $r$ of type $(a,b)$ where $\min(a,b)=1$.

Denote by $\mathbb{P}_{k}$ the set of $k$-th powers of primes and suppose $b>a$, so that $(n-r,n+r)=(p,q^{b})$ with $p$ and $q$ odd primes. Provided $n>5^{b}$, consider the closest odd element $q'$ of $\mathbb{P}_{b}$ to $p$ less than $p$ and set $2t=p-q'$. Now consider the closest prime $p'$ to $n+r+2t$ greater than $n+2t$. Then $2r':=p'-q'$ is twice a Galois radius of $n$ of type $(b,a)$ and we may set $s:=\vert r-r'\vert$, which that way fulfills the required conditions to be an inversion shift of $r$. The same line of reasoning applies mutatis mutandis for the case $b<a$.

Only a partial answer for the special case of an integer $n$ with Galois radius $r$ of type $(a,b)$ where $\min(a,b)=1$.

Denote by $\mathbb{P}_{k}$ the set of $k$-th powers of primes and suppose $b>a$, so that $(n-r,n+r)=(p,q^{b})$ with $p$ and $q$ odd primes. Provided $n>5^{b}$, consider the closest odd element $q'$ of $\mathbb{P}_{b}$ to $p$ less than $p$ and set $2t=p-q'$. Now consider the closest prime $p'$ to $n+r+2t$ greater than $n+r+2t$. Then $2r':=p'-q'$ is twice a Galois radius of $n$ of type $(b,a)$ and we may set $s:=\vert r-r'\vert$, which that way fulfills the required conditions to be an inversion shift of $r$. The same line of reasoning applies mutatis mutandis for the case $b<a$.