Thread: mod121

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

$\displaystyle 121= 11^2$. Obviously $\displaystyle (121)^2+ 3(121)+ 5$ is congruent to 5 (mod 121) and any other number that divides 121 must also divide 11. So it is sufficient to do an "exhaustive search" of numbers in [0, 10].