Solve in Z : $\displaystyle x^{2}+3x+5\equiv 0 (mod121)$
Hello, Dhiab,
Solution : $\displaystyle x^2 + 3x + 5$ never divides $\displaystyle 121$, so this equation has no solutions in $\displaystyle \mathbb{Z}$
Proof : $\displaystyle x^2 + 3x + 5 \equiv (x \pmod{121})^2 + 3(x \pmod{121}) + 5 \pmod{121}$. Apply an exhaustive search on $\displaystyle x \in [0,120]$ (since values wrap around beyond this interval), and prove that none of the values of the quadratic modulo $\displaystyle 121$ are divisible by $\displaystyle 121$.
Sorry, didn't come up with anything smarter yet![]()