2$. In that case $M_{b,q}$ is a quadratic residue modulo $p$ and $p\in P_b$. Now if $p>2$ and $p\mid (b-1)$, then $M_{b,q}\equiv q$ (mod $p$). By Dirichlet's theorem, we know there are infinitely many primes $q$ in an arbitrary reduced residue class modulo $p$, in particular the class of a quadratic non-residue; hence $p\not\in M_b$ in this case. The second part of the claim follows as $2\mid b(b+1)$ and that $\{b-1,b,b+1\}$ is a complete residue system modulo 3. \end{proof} \begin{remark}\label{Pb>L} In particular, $P_b$ contains all the prime factors of $b$. Hence, given an arbitrarily large number $L$, we can find a base $b$ such that $|P_b|\geq L$, e.g., by considering $b$ which is the product of $L$ distinct primes. \end{remark} \begin{remark} It is clear from the definition that we have $p\in P_b$ if and only if $p\in P_B$ for all $B\equiv b$ (mod $p$). Hence, for a fixed prime $p$, we can identify all the bases $b$ for which $p\in P_b$ once they have been determined up to $b

r_1>r_2>\cdots >r_t>1$. With these, we apply the reciprocal law to the Jacobi symbol $\leg{b^m-1}{b^{2^n}+1}$ repeatedly, $$ \dleg{b^{2^s r}-1}{b^{2^n}+1}= \dleg{b^{2^n}+1}{b^{2^s r}-1}= \dleg{b^{2^s r_1}+1}{b^{2^s r}-1}= \dleg{b^{2^s r}-1}{b^{2^s r_1}+1}= \dleg{(-1)^{u_1}b^{2^s r_2}-1}{b^{2^s r_1}+1} $$ Note that $\leg{-1}{b^{2^s r_1}+1}=+1$. Depending on the parity of $u_1$, we continue, $$ \dleg{(-1)^{u_1}b^{2^s r_2}-1}{b^{2^s r_1}+1}= \dleg{\pm b^{2^s r_2}-1}{b^{2^s r_1}+1}= \dleg{b^{2^s r_2}\mp 1}{b^{2^s r_1}+1}= \dleg{b^{2^s r_1}+1}{b^{2^s r_2}\mp 1}. $$ At this point, there are three ways to go with the $+/-$ sign: $$ \dleg{b^{2^s r_1}+1}{b^{2^s r_2}\mp 1}= \dleg{b^{2^s r_3}+1}{b^{2^s r_2}-1} \quad\text{or}\quad \dleg{b^{2^s r_3}+1}{b^{2^s r_2}+1} \quad\text{or}\quad \dleg{b^{2^s r_3}-1}{b^{2^s r_2}+1}. $$ In fact, these three are the only possible forms until we reach the $t$-th iteration, i.e., $$ \dleg{b^{2^s r}-1}{b^{2^n}+1}= \dleg{b^{2^s}+1}{b^{2^s r_t}-1} \quad\text{or}\quad \dleg{b^{2^s}+1}{b^{2^s r_t}+1} \quad\text{or}\quad \dleg{b^{2^s}-1}{b^{2^s r_t}+1}. $$ We proceed case by case, respectively: \medskip \noindent\emph{Case 1.} Note that here $r_t$ must be odd since we start off with the Jacobi symbol $\leg{b^m-1}{b^{2^n}+1}$ having $\gcd(b^m-1,b^{2^n}+1)=1$. Hence, $$ \dleg{b^{2^s}+1}{b^{2^s r_t}-1}= \dleg{b^{2^s r_t}-1}{b^{2^s}+1}= \dleg{(-1)^{r_t}-1}{b^{2^s}+1}= \dleg{-2}{b^{2^s}+1}. $$ \emph{Case 2.} This time, $r_t$ is even: $$ \dleg{b^{2^s}+1}{b^{2^s r_t}+1}= \dleg{b^{2^s r_t}+1}{b^{2^s}+1}= \dleg{(-1)^{r_t}+1}{b^{2^s}+1}= \dleg{2}{b^{2^s}+1}. $$ \emph{Case 3.} Similarly, $$ \dleg{b^{2^s}-1}{b^{2^s r_t}+1}= \dleg{b^{2^s r_t}+1}{b^{2^s}-1}= \dleg{2}{b^{2^s}-1}. $$ From all these we collect the following results. \begin{enumerate} \item For $s=0$, we have $\leg{b^m-1}{b^{2^n}+1}=+1$ if $b\equiv 0$ (mod 8), and $\leg{b^m-1}{b^{2^n}+1}=-1$ if $b\equiv 4$ (mod 8), and indeterminate if $b\equiv 2$ (mod 4). \item For $s=1$, we have $\leg{b^m-1}{b^{2^n}+1}=+1$ if $b\equiv 0$ (mod 4), and $\leg{b^m-1}{b^{2^n}+1}=-1$ if $b\equiv 2$ (mod 4). \item For $s\geq 2$, we have $\leg{b^m-1}{b^{2^n}+1}=+1$ in all cases. \end{enumerate} We also verify easily that $\leg{b-1}{b^{2^n}+1}=+1$ if $b\equiv 0,2$ (mod 8), and $\leg{b-1}{b^{2^n}+1}=-1$ if $b\equiv 2,4$ (mod 8). Lastly, we use the relation $\leg{M_{b,m}}{F_{b,n}}=\leg{b^m-1}{b^{2^n}+1}\leg{b-1}{b^{2^n}+1}$ to put together the desired claim. \end{proof} \begin{corollary}\label{b04} Suppose that $b\equiv 0\pmod 4$. Then $\leg{M_{b,q}}{F_{b,n}}=+1$ for any odd number $q$. Hence, if $F_{b,n}$ is prime, then $F_{b,n}\in P_b$. In reciprocal, if $M_{b,q}$ is prime, then $M_{b,q}$ is anti-elite to the base $b$. \end{corollary} \begin{remark} This corollary is the case $s=0$ in Theorem \ref{Fbn}. Primes $M_{b,q}$, being scarce, are typically discovered having large values of $q$. Now, since Fermat numbers are recursive, they are periodic modulo $p$. In the case of $p=M_{b,q}$, the period length is given by $|2|_q$, hence expectedly large too. For instance, the fourteenth repunit prime base 12, according to \seqnum{A004064} in OEIS \cite{oeis}, is $M_{12,769543}$. Since 12 is a multiple of 4, this prime is anti-elite to this base with period $|2|_{769543}=384771$. \end{remark} \begin{corollary}\label{FnP4} Suppose that $F_{b,n}$ is a prime number for some $n\geq 1$. If $b\equiv 0\pmod 4$, then $F_{b,n}\in P_{b^k}$ for all $k\geq 1$ such that $2^{n+1}\ndiv k$. If $b\equiv 2\pmod 4$, then $F_{b,n}\in P_{b^{2k}}$ for all $k\geq 1$ such that $2^{n}\ndiv k$. In particular, all Fermat primes $F_n\in P_{4^k}$ for all such $k$. \end{corollary} \begin{proof} The identity $M_{b,km}=M_{b^k,m}M_{b,k}$ gives $\leg{M_{b^k,m}}{F_{b,n}}=\leg{M_{b,km}}{F_{b,n}}\leg{M_{b,k}}{F_{b,n}}$. Note that if $m$ is odd, the factor of 2 dividing $k$ and $km$ are of the same multiplicity. In that case $\leg{M_{b,km}}{F_{b,n}}=\leg{M_{b,k}}{F_{b,n}}$ by Theorem \ref{Fbn}, provided that we require $k$ to be even when $b\equiv 2$ (mod 4) in order to avoid the indeterminate case. Hence, $\leg{M_{b^k,q}}{F_{b,n}}=+1$ for all odd prime $q$. \end{proof} \begin{remark} We can extend the proof of Theorem \ref{Fbn} to include $\leg{M_{b^k,m}}{F_{b,n}}=+1$ if $m$ is even, provided that $k$ is also even, or a multiple of 4 when $b\equiv 2$ (mod $4$). In particular, with $m=2$ and $k=2^r$, we have $M_{b^{2^r},\,2}=F_{b,r}$. Then $\leg{F_{b,n}}{F_{b,r}}=\leg{F_{b,r}}{F_{b,n}}=+1$ for all $n>r$. Hence, we can show that generalized Fermat primes $F_{b,r}$, for $r\geq 2$, are anti-elite to their own base. However, this is a fact already known \cite{mul} and easy to obtain due to the relation $F_{b,n}\equiv 2$ (mod $F_{b,r}$) for all $n>r$. And unfortunately, no other anti-elite primes can be identified in this manner as it has been observed that $M_{b,q}$ is always composite when $b$ is a perfect power, except possibly for $q=2$, i.e., the Fermat primes already considered. \end{remark} We do not observe a general pattern for odd bases $b$. The following theorem deals with an isolated result for $b=3$. \begin{theorem} If the number $\frac{F_{3,n}}{2}$ is prime, then $\frac{F_{3,n}}{2}\in P_{9^k}$ for all $k$ odd. \end{theorem} Primes of the form $\frac{F_{3,n}}{2}$ are currently known with $n=0,1,2,4,5,6$ [\seqnum{A275377}]. Four of these are included in Table \ref{one}, belonging in $P_9$. \begin{proof} We exclude $n=0$, although it is valid, and introduce the notation $E_r=\frac{9^r+1}{2}$. Hence $\frac{F_{3,n}}{2}=E_{2^{n-1}}$. Note the following facts. \begin{enumerate} \item We have $E_r\equiv 1$ (mod 4) for all $r\geq 0$. \item The number $M_{9,m}$ is odd only when $m$ is odd. \item We have $\gcd(E_{2^{n-1}},M_{9,m})=1$ for all $n\geq 1$ when $m$ is odd. \end{enumerate} We first show that $\leg{M_{9,m}}{E_{2^{n-1}}}=+1$ for all odd $m$, then it would follow that $\leg{M_{9^k,m}}{E_{2^{n-1}}}=\leg{M_{9,km}}{E_{2^{n-1}}}\leg{M_{9,k}}{E_{2^{n-1}}}=(+1)(+1)=+1$ for all odd $k$. Let us fix an odd $m$, and let $r_1=2^{n-1}\bmod m$. By the congruence $9^m\equiv 1$ (mod $M_{9,m}$), we get $$ \dleg{E_{2^{n-1}}}{M_{9,m}}= \dleg{2}{M_{9,m}}\dleg{9^{2^{n-1}}+1}{M_{9,m}}= \dleg{2}{M_{9,m}}\dleg{9^{r_1}+1}{M_{9,m}}= \dleg{E_{r_1}}{M_{9,m}}. $$ Now by the congruence $9^{r_1}\equiv -1$ (mod $E_{r_1}$), this Jacobi symbol equals to $$ \dleg{M_{9,m}}{E_{r_1}}= \dleg{8}{E_{r_1}}\dleg{9^m-1}{E_{r_1}}= \dleg{8}{E_{r_1}}\dleg{\pm 9^{r_2}-1}{E_{r_1}}, $$ where $r_2=m\bmod r_1$. In the plus case, we have $$ \dleg{E_{2^{n-1}}}{M_{9,m}}= \dleg{8}{E_{r_1}}\dleg{9^{r_2}-1}{E_{r_1}}= \dleg{M_{9,r_2}}{E_{r_1}}= \dleg{E_{r_1}}{M_{9,r_2}}, $$ noting that if $r_2$ is even, then the Jacobi symbol is temporarily replaced by Kronecker symbol. Similarly, in the minus case, $$ \dleg{E_{2^{n-1}}}{M_{9,m}}= \dleg{8}{E_{r_1}}\dleg{-9^{r_2}-1}{E_{r_1}}= \dleg{2}{E_{r_1}}\dleg{9^{r_2}+1}{E_{r_1}}= \dleg{E_{r_2}}{E_{r_1}}= \dleg{E_{r_1}}{E_{r_2}}. $$ This process terminates, say at the $t$-th iteration, having produced the remainders $r_1>r_2>\cdots>r_t>1$. Observe that there are only three possible final forms. \noindent \emph{Case 1.} We have $\leg{E_{2^{n-1}}}{M_{9,m}}=\cdots=\leg{M_{9,1}}{E_{r_t}}=+1$. \noindent \emph{Case 2.} The case where we must have $r_t$ odd, since $\gcd(E_{2^{n-1}},M_{9,m})=1$: $$ \dleg{E_{2^{n-1}}}{M_{9,m}}= \cdots= \dleg{E_1}{M_{9,r_t}}= \dleg{M_{9,r_t}}{5}= \dleg{8}{5}\dleg{9^{r_t}-1}{5}= -\dleg{-2}{5}=+1. $$ \noindent \emph{Case 3.} The last case, where we would have $r_t$ even: $$ \dleg{E_{2^{n-1}}}{M_{9,m}}= \cdots= \dleg{E_1}{E_{r_t}}= \dleg{5}{E_{r_t}}= \dleg{2}{5}\dleg{9^{r_t}+1}{5}= \dleg{2}{5}\dleg{2}{5}=+1. $$ And the proof is complete. \end{proof} Our last observation in connection with anti-elite primes is irrespective of the family to which the prime $p$ belongs, but is given by the multiplicative order of the base $b$. \begin{theorem}\label{oddord} Suppose that $|b|_p$ is an odd number. If $p\in P_b$, then $p$ is anti-elite to the base $b$. \end{theorem} Table \ref{one} contains quite a number of occurrences with odd $|b|_p$, thus anti-elite primes, e.g., 2906161 in base 10 and 2413941289 in base 3, as well as the repunit $M_{3,13}=797161\in P_3$. \begin{proof} Note that $F_{b,n}\,M_{b,2^n}=M_{b,2^{n+1}}$ and that $\gcd(2^n,|b|_p)=1$ for all $n$. Hence, if $p\in P_b$, then by Theorem \ref{alg}, both $M_{b,2^n}$ and $M_{b,2^{n+1}}$ are quadratic residues modulo $p$, and so is $F_{b,n}$. \end{proof} \begin{corollary}\label{b34} Every repunit prime $M_{b,q}$ is anti-elite to its base, if $b\equiv 0, 3\pmod 4$. \end{corollary} \begin{proof} The case $b\bmod 4=0$ overlaps with Corollary \ref{b04}. Regardless, for these two classes of $b$, we have $M_{b,q}\in P_b$ if prime, by Theorem \ref{Mbq}. Such an event comes with $|b|_{M_{b,q}}=q$, which is necessarily an odd prime. \end{proof} \section{For further research} This has been an initial investigation on primes $p$ elements of $P_b$, i.e., such that $\leg{M_{b,q}}{p}=+1$ for all large primes $q$, and motivated by the search for new insight into elite/anti-elite primes. If the subject is proven worthwhile, we would invite the readers to look into the infinitude of $P_b$, for all or specific values of $b$. The facts that some $P_b$ can be large (Remark \ref{Pb>L}) and that some contain repunit primes naturally suggest that there may indeed be infinitely many such primes for a fixed base. However, our numerical data also suggest that we may have a convergent reciprocal sum $\sum_{p\in P_b}1/p$, in line with the reciprocal sum of elite primes, which has been proved convergent by K\v r\'i\v zek et al.~\cite{kls}. \begin{thebibliography}{9} \bibitem{aig} A. Aigner, \"Uber Primzahlen, nach denen (fast) alle Fermatschen Zahlen quadratische Nichtreste sind, {\it Monatsh. Math.} {\bf 101} (1986), 85--93. \bibitem{dub} H. Dubner, Generalized repunit primes, {\it Math. Comp.} \textbf{61} (1993), 927--930. \bibitem{kls} M. K\v r\'i\v zek, F. Luca, and L. Somer, On the convergence of series of reciprocals of primes related to the Fermat numbers, {\it J. Number Theory} {\bf 97} (2002), 95--112. \bibitem{leb} M. Lebesgue, Sur l'impossibilit\'e, en nombres entiers, de l'\'equation $x^m=y^2+1$, {\it Nouvelles Annales de Math\'ematiques 1\textsuperscript{re} s\'erie} {\bf 9} (1850), 178--181. \bibitem{mul} T. M\"uller, On anti-elite prime numbers, {\it J. Integer Sequences} {\bf 10} (2007), \href{https://cs.uwaterloo.ca/journals/JIS/VOL10/Mueller/mueller56.html}{Article 07.9.4}. \bibitem{oeis} N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, \url{https://oeis.org}, last accessed January 2021. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2020 {\it Mathematics Subject Classification}: Primary 11A15; Secondary 11A41. \noindent \emph{Keywords: } Fermat number, Mersenne number, generalized repunit, elite prime. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A004064}, \seqnum{A102742}, \seqnum{A128852}, and \seqnum{A275377}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received August 4 2020; revised version received January 2 2021. Published in {\it Journal of Integer Sequences}, January 3 2021. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}