\documentclass[11pt]{amsart} \pagestyle{empty} \title{Comment on Solution of Quadratic Congruences} \author{Alberto Tonelli} \thanks{ Cf.~\emph{Bemerkung \"uber die Aufl\"osung quadratischer Congruenzen}, Nachr.~d.~Akad. d. Wiss. in G\"ottingen 1891, 344--346. Excerpt from letters dated 1891-04-18 and 1891-06-15.} \begin{document} \begin{abstract} We have extended the well-known method of general solution of a quadratic congruence for prime moduli not of the form $4h+1$, in the way that it can now be applied to primes of said form. \end{abstract} \maketitle If the congruence $$xx\equiv c \pmod{p}$$ is to be solved, and also one arbitrary quadratic non-residue $g\pmod{p}$ is known, then the following shows the algorithm. Let $p=\alpha2^s+1$, where $\alpha$ is odd and $s\ge1$, then it holds by the theorem of Euler that $$c^{\alpha2^{s-1}}\equiv+1,\quad g^{\alpha2^{s-1}}\equiv-1\pmod{p}.$$ If also $s\ge2$ then from the first of both congruences a new one follows $$c^{\alpha2^{s-1}}\equiv\pm1\pmod{p}.$$ Let now $\varepsilon_0=0$ if the upper sign $(+)$ applies, and $\varepsilon_0=1$\footnote{['$\varepsilon_1=1$' in the original --- RS]} if it is the lower $(-)$, so it always holds that $$g^{\varepsilon_0\alpha2^{s-1}}c^{\alpha2^{s-2}}\equiv\pm1\pmod{p}.$$ If now additionally $s\ge3$ then follows $$g^{\varepsilon_0\alpha2^{s-2}}c^{\alpha2^{s-3}}\equiv\pm1\pmod{p}.$$ If $(+)$ holds here then we set $\varepsilon_1=0$, and if it is $(-)$ then $\varepsilon_1=1$, so that always $$g^{\varepsilon_1\alpha2^{s-1}}g^{\varepsilon_0\alpha2^{s-2}}c^{\alpha2^{s-3}}\equiv\pm1\pmod{p}.$$ This way one gets the congruence $$g^{\alpha2^{s-k}(\varepsilon_0+\varepsilon_12+\dotsb+\varepsilon_{k-1} 2^{k-1})}c^{\alpha2^{s-k-1}}\equiv1\pmod{p}$$ as long as $k