Does $$x^2+5x \equiv 12 \pmod{31} $$ have a solution?
2026-03-26 04:31:58.1774499518
Quadratic congruence $x^2+5x \equiv 12 \pmod{31} $
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The equation is equivalent to: $$ 4x^2 + 20x\equiv 48 \equiv 17\pmod{31}$$ or to: $$ (2x+5)^2 \equiv 11\pmod{31} $$ hence it has two solutions iff $11$ is a quadratic residue $\!\!\pmod{31}$. So, let us compute the Legendre symbol $\left(\frac{11}{31}\right)$ through quadratic reciprocity: $$\left(\frac{11}{31}\right) = - \left(\frac{31}{11}\right) = -\left(\frac{9}{11}\right) = -1.$$ No way, $11$ is not a quadratic residue $\!\!\pmod{31}$, hence there are no solutions.