Solving $x^2 + 7x \equiv 1\pmod{13}$

Question

How should I solve this particular congruence: $x^2 + 7x \equiv 1 \pmod{13}$? I can re-arrange the equation to get $$x^2 + 7x - 1 \equiv_{13} 0$$. In noticing that $7 \equiv_{13} -6$, can I replace $7x$ by $-6x$ in order that the middle term should be preceded by an even coefficient?

Does it help you to do that? What if you noticed $-1 \equiv_{13} 12$? — Osama Ghani, Sep 24 '23 at 22:43
\pmod{13} automatically gives you the space, the typeface, and the parentheses. — Arturo Magidin, Sep 24 '23 at 23:01
As in the dupe, the quadratic formula (completing the square) works over any field. Here the discriminant is $53\equiv 1$ so the roots are $,(-7\pm 1)/2 \equiv (6\pm 14)/2\equiv 3\pm 7\ \ $ — Bill Dubuque, Sep 24 '23 at 23:24

score 1 · Accepted Answer · answered Sep 24 '23 at 22:49

1

Note that $x^2 + 7x - 1 \equiv_{13} 0$ is the same as $x^2 + 7x + 12 \equiv_{13} 0$. Now note that the left hand side factors as $(x+4)(x+3)$. So, $x = -3,-4$ i.e. $x = 9,10$.

answered Sep 24 '23 at 22:49

SescoMath

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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Sep 24 '23 at 23:32

Solving $x^2 + 7x \equiv 1\pmod{13}$

1 Answers1