For which prime p does the congruence (x^2 + 4x + 5) mod p =0 have
a solution? some hint about this.
1) if p = 2 then $\displaystyle x^2+4x+5=x^2+1=(x+1)^2\!\!\!\pmod 2$ and we have a double root;
2) If p > 2 then the equation has a solution iff the quadratic's determinant is a square modulo p, i.e. $\displaystyle \Delta=16-20=-4$ is a square modulo p $\displaystyle \Longleftrightarrow -1 $ is a square modulo p.
For example, for p = 5,13,17,29, 97 we have a solution, pero not for p = 3, 7, 19, 31
Tonio