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Given $x^4 + 36x^3 - 19x^2 + 11x - 14 \equiv 0 \pmod{5}$. How would one go about solving such an congruence equation? Maybe it's possible to reduce this to a quadratic congruence? I can't figure it out! Any help is appreciated.

jazzinsilhouette
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2 Answers2

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Hint $\ {\rm mod}\ p\!:\ f \equiv x^{p-1}\!+\,\cdots+\!x^2\!+\!x\!+\!1\equiv 0\,\Rightarrow\, (x\!-\!1)f \equiv \color{#c00}{x^p}\!-1\equiv \color{#c00}x\!-\!1\equiv 0\ $ by $\,\rm\color{#c00}{little\ Fermat}$.

Bill Dubuque
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  • Thanks, Bill. This is really quite neat. From this it's obvious that $x = 1$ is the only solution. – jazzinsilhouette Jan 25 '15 at 16:13
  • @Number Any chance you can do that again using this equation instead? $x^4 + 21x^3 + 5x^2 + 7x + 1 \equiv 0 \mod 23$ I know the answer is x=18 | x=19 but I don't know the steps involved nor do I get how you're only using 1s in your answer when the OP had integers > 1. I suspect it has something to do with their example, so I'm hoping the theorem applied to my example will shed some light since the solution to my equation is not 1. – Levitikon Jul 29 '18 at 19:48
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Your equation is equivalent to $$x^4+x^3+x^2+x+1\equiv 0\pmod{5}$$ since $36\equiv -19\equiv 11\equiv -14\equiv1\pmod{5}$ You can test all the numbers $0,1,2,3,4$ and see that the only one that fits is $1$.

kingW3
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  • Thanks a lot! Can you maybe point out how -19 can be seen to be congruent 1 mod 5? I see that it's just consecutively adding fives to -19, but what's the formal background? I'm still a little confused when it comes to negative integers in modular arithmetic, after all there're no "negative congruence classes", so they have to end up in 0,1,...,p-1 – jazzinsilhouette Jan 25 '15 at 15:22
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    No problem,well you have that $-19=-20+1$ which means that it's equivalent to $1$ since $20$ is a multiply of 5 other then that you have that $-19\equiv -4\pmod{5}$.We can add 5 since it's equivalent to adding a zero hence adding $5$ we get that $-4\equiv 5-4\equiv 1\pmod{5}$ – kingW3 Jan 25 '15 at 15:26
  • I see. Thanks again. – jazzinsilhouette Jan 25 '15 at 15:27
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    @jazz You can eliminate the brute-force checking by using little Fermat, obtaining a much more general result - see my answer. – Bill Dubuque Jan 25 '15 at 15:59