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Problem :

I'm going today to find all natural number such that : $1163 \mid 3n^{3}+2$ Mathematical give $n \equiv 435 \mod 1163$

My little try :

first we note that $1163$ prime number

$3n^{3}\equiv -2 \pmod{1163}$

Now I don't know how I complete my work My be we can use Fermat's little theorem

Daniel Fischer
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Ellen Ellen
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2 Answers2

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The first step is to find that $3^{-1} \equiv 388 \pmod{1163}$ (can be done with Euclid's algorithm). So $$ 3n^3 + 2 \equiv 0 \iff n^3 \equiv -2\cdot 388 \equiv 387 \pmod{1163} $$ We know that $(\mathbb Z/1163\mathbb Z)^\times$ is cyclic, because $1163$ is prime. If $(\mathbb Z/1163\mathbb Z)^\times = \langle x \rangle$, then $x^3$ also generates the group, because $3$ and $1162$ (the order of the group) are coprime. That means that $a\mapsto a^3$ is an automorphism (in particular a bijection) of $(\mathbb Z/1163\mathbb Z)^\times$. In other words, any equation $n^3 \equiv m \pmod{1163}$ has a unique solution.

Now we need to find the solution. Note that $3^{-1} \equiv 775\pmod{1162}$. Thus: $$ n^3 \equiv 387 \equiv 387^{3\cdot 775} \pmod{1163} $$ Which gives $n\equiv 387^{775} \equiv 435 \pmod{1163}$.

EDIT: How to find $3^{-1}$ (with a general approach).

Use the Euclidean algorithm on $(1163,3)$: $$ 1163 = 387 \cdot 3 + 2 \\ 3 = 1 \cdot 2 + 1 $$ Now we work backwards: $$ \begin{align} 1 &= 3 - 1\cdot 2 \\ &= 3 - 1 \cdot (1163 - 387\cdot 3) \\ &= 388\cdot 3 - 1\cdot 1163 \end{align} $$ We see that $388\cdot 3 \equiv 1\pmod{1163}$, so $3^{-1} \equiv 388 \pmod{1163}$. The same approach can be used to find $3^{-1} \equiv 775\pmod{1162}$.

Milten
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  • Thanks sir @Milten , but how you get $\frac{1}{3}\equiv 388\pmod{1163}$ – Ellen Ellen Nov 01 '19 at 11:47
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    or note -2 is 1161, leading to $n^3\equiv 387\equiv -776$ neither of which is a cube normally so we can eliminate -9 to 7 from possibilities. –  Nov 01 '19 at 11:47
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    @EllenEllen $1164/3= 388$ –  Nov 01 '19 at 11:48
  • Ooh , OK @Macphee thank you very much , actually I don't understand all the solution – Ellen Ellen Nov 01 '19 at 11:51
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    @EllenEllen Both of Roddy MacPhee's suggestions are good. For a general approach, see here. I believe the Euclidean algorithm is usually preferred. There are also a bunch of questions on this site about it. – Milten Nov 01 '19 at 11:56
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    @EllenEllen You can also find it as $3^{-1}\equiv 3^{1162-1}$, but that is quite inefficient to compute. – Milten Nov 01 '19 at 11:58
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    @EllenEllen In short: Use the Euclidean algorithm on $(1163, 3)$ until you get to the last line $r_k = r_{k+1}+1$. Then you can substitute backwards in your equations to write $3x+1163y=1$, which means $3^{-1} \equiv x$. – Milten Nov 01 '19 at 12:02
  • Let me see @Milten , then I tell you if I get it – Ellen Ellen Nov 01 '19 at 12:04
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Hint $\,\ $ $3^{\large 7}\! = 2^{\large 10}\! + 1163\,$ $\Rightarrow\, \bmod 1163\!:\ n^{\large 3}\equiv {-}\dfrac{2}3 \equiv \left[\dfrac{-9}8\right]^{\large 3}$ and $\ \, \dfrac{-9}8 \equiv\, \bbox[5px,border:1px solid #c00]{\!\!435}$

Bill Dubuque
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