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Determine all the solutions of the congruence $5x^{44} ≡ 9 \pmod{29}$, using the index in base $2$ module $29$.

I don't know how to proceed with the presence of this $5$ by multiplying the $x$ ... should I multiply by the inverse class to eliminate it?

gmn_1450
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  • Yes, you should multiply by inverse of $5\pmod{29}$. You will end up with $x^{44}\equiv\text{something}\pmod{29}$. –  Dec 03 '20 at 13:34
  • One also has the identity that $x \equiv x^{29} \pmod{29}$ which can be used to simplify the question further to $x^{15} \equiv \text{something} \pmod{29}$. – Aryaman Maithani Dec 03 '20 at 13:41
  • @Aryaman That (and more) is already incorporated (implciitly) by "index calculus", i.e. by the properties of $,\ell_2,$ = discrete log to base $2,,$ i.e. $\bmod 29!:\ 2,$ has order $= \color{#c00}{28},$ so by mod order reduction $$\large \begin{align}2^a\equiv 2^b!!!\pmod{!29}!&\iff\ \ \ \ \ a^{\phantom{|^|}}!\equiv\ b\ \ \ \ \ \ , \pmod{!\color{#c00}{28}}\[.2em] {\rm i.e.}\ \ \ A\equiv B!!\pmod{!29}!&\iff! \ell_2(A)\equiv \ell_2(B)!!\pmod{!\color{#c00}{28}} \end{align}\qquad$$ I use this viewpoint in my answer below. – Bill Dubuque Dec 03 '20 at 15:56

2 Answers2

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Yes, $\bmod 29\!:\ \color{#0a0}{\dfrac{9}5}\equiv \dfrac{54}{30}\equiv \dfrac{(\color{#c00}{-1})4}1\equiv \color{#c00}{2^{\large 14}}2^{\large 2}\equiv \color{#0a0}{2^{\large 16}},\,$ so with $\,x = 2^{\large n}\,$ we have

$ x^{\large 44}\!\equiv \color{#0a0}{\dfrac{9}5}\!\!\iff\!\!2^{\large 44n}\!\equiv \color{#0a0}{2^{\large 16}}\!\!\!\iff\! \bmod 28\!:\, \underbrace{44n}_{\textstyle 16n}\equiv 16\!\!\!\underset{\large \div\ 4\!\!}\iff\! \bmod 7\!:\ 4n\equiv 4\!\!\iff\! n\equiv 1$

Bill Dubuque
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$$2^5\equiv3\pmod{29}\implies9\equiv(2^5)^2$$

$2^{14}=(2^5)^22^4\equiv3^22^4\equiv-1\pmod{29}$

So, $2$ is a primitive root $\pmod{29}$

$$\implies5\equiv-2^3\cdot3\equiv2^{14+3+5}$$

Using Discrete logarithm with index in base $2,$

$22+44$ind$_2x\equiv10\pmod{\phi(29)}$

$\iff44$ind$_2x\equiv10-22\pmod{\phi(29)}\equiv-12+28\pmod{28}$

$\iff11$ind$_2x\equiv4\pmod7$

$\iff4$ind$_2x\equiv4\pmod7$ as $(11\equiv4\pmod7)$

$\iff$ind$_2x\equiv1\pmod7$ as $(4,7)=1$

$\implies x\equiv2^{1+7k}\pmod{29}, 0\le1+7k\le28\iff0\le k\le3$

lab bhattacharjee
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