Past exam of the options II course of the University of Oxford / 2012 / C / C2 / c / Solution

Reducing modulo $23$ gives $189\equiv 5$. Reciprocity with five has positive sign, sosince the nonzero squares modulo five are one and four. Modulo $59$, $189\equiv 12=4\cdot 3$, and $59\equiv -1(\mathrm{mod}12)$. Thus part (b) givesAs $1357=23\cdot 59$, a solution modulo $1357$ would in particular solve the modular congruence modulo $23$, contradicting the first symbol. Therefore there is no integer solution modulo $1357$.