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I have this question but I am not sure how to proceed on it.

Find all values $a$ for which the system of the two equations $$xy=a, \quad x+y=1$$ has a solution in $\mathbb{Z}/19\mathbb{Z}$. Is there such an $a$ for which there is a unique solution?

I am not sure if the $a$ has to be in $\mathbb{Z}/19\mathbb{Z}$ or if the solutions are, so far I have $x-x^2=a$ and $y=1-x$.

Thanks for your help.

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I'm not sure what you mean when you ask if $a$ has to be in $\mathbb{Z}/(19)$, since it doesn't really matter if $a$ is in $\mathbb{Z}$ or $\mathbb{Z}/(19)$ since the equation is mod 19 regardless.

Now note that since you have $x-x^2=a$, then we can complete the square to get $x^2-x + 5 = (x-10)^2 = 5-a$. Thus we need $$ \left(\frac{5-a}{19}\right)=1$$ in order to have two solutions, or $a=5$ to have a single unique solution of $x=10$.

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  • $\begingroup$ First of all thanks a lot for your answer. Now I am not sure I understand properly why you take the (x-10)^2 and where it comes from ? (x-10)^2 is not equal to x^2-x+5, is it ? Or is it because it's mod 19 ? $\endgroup$
    – TedMosby
    Oct 26, 2015 at 5:01
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    $\begingroup$ @ThéodoreRozencwajg Yep, it's because it's mod 19, $100\equiv 5 \pmod{19}$, and $-20 \equiv -1 \pmod{19}$ as well. Also it just comes from the usual completing the square, $\frac{-1}{2}\equiv -10 \pmod{19}$. $\endgroup$
    – jgon
    Oct 26, 2015 at 5:02
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    $\begingroup$ @ThéodoreRozencwajg It depends on your convention. I tend to use equals signs when working mod a certain number and the number doesn't change, but technically the congruences are more proper. So (mod 19) there are 9 choices for $a$ that have 2 solutions for $x$ each (where $5-a$ is a nonzero perfect square modulo 19), and one solution for $a$ which has a unique solution for $x$ (when $a=5\equiv -14\pmod{19}$). $\endgroup$
    – jgon
    Oct 26, 2015 at 5:11
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    $\begingroup$ So that's not a fraction, I should explain, sorry. $\endgroup$
    – jgon
    Oct 26, 2015 at 21:08
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    $\begingroup$ That's the Legendre Symbol: wikiwand.com/en/Legendre_symbol. $\endgroup$
    – jgon
    Oct 26, 2015 at 21:09

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