mod121

**dhiab** · Feb 25th 2010, 12:16 AM

Solve in Z : $x^{2}+3x+5\equiv 0 (mod121)$

**Bacterius** · Feb 25th 2010, 01:07 AM

Hello, Dhiab,

Solution : $x^2 + 3x + 5$ never divides $121$ , so this equation has no solutions in $\mathbb{Z}$

Proof : $x^2 + 3x + 5 \equiv (x \pmod{121})^2 + 3(x \pmod{121}) + 5 \pmod{121}$ . Apply an exhaustive search on $x \in [0,120]$ (since values wrap around beyond this interval), and prove that none of the values of the quadratic modulo $121$ are divisible by $121$ .

Sorry, didn't come up with anything smarter yet

**HallsofIvy** · Feb 25th 2010, 01:20 AM

$121= 11^2$ . Obviously $(121)^2+ 3(121)+ 5$ 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].

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