p3sa1bynn

2022-04-27

Solve ${x}^{2}+3x+17\equiv 0\pm \mathrm{mod}315$

otposlati9u8

Beginner2022-04-28Added 19 answers

${x}^{2}+3x+17\equiv 0\pm \mathrm{mod}(5,7,9)$

can only have 5,7 or 9 solutions. Try all of them.

Alternately:

${x}^{2}+3x+17\equiv 0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}315\right)\Rightarrow $

$4{x}^{2}+12x+68\equiv 0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}315\right)\Rightarrow $

$(2x+3{)}^{2}\equiv 256={16}^{2}\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}315)$

You can immediatelly see that 16,-16 are square roots of 256, which immediately tells you that there are at least 2 solutions. Since 3,5,7 do not divide 256 it is easy to deduce that there must be 8 solutions, you just need to find the rest.

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