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To solve the congruence $$ 9^x \equiv 4 \pmod{17}$$, I thought of multiplying both sides by (13), the modular inverse of ( 4 ) modulo ( 17 ).

Since $$ 13 \equiv 9^2 \pmod{17}.$$ , I rewrote the congruence as:

$$(9^3)^x \equiv 1 \pmod{17}.$$

From here, I know the order of $$ 9 \pmod{17} $$ is $$ 8 $$, so the condition for $$ x $$ is that $$ 3x \equiv 0 \pmod{8}$$. If the reasoning is correct, can I conclude with x=8k, where k is an integer.

Bill Dubuque
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Irene
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    How did you get $(9^3)^x \equiv 1$? – mihaild Jan 06 '25 at 14:28
  • You can check your own answer and $k=0$ fails the test. – Marius S.L. Jan 06 '25 at 14:33
  • You are right, but than I do not understand ho to approach this problem – Irene Jan 06 '25 at 14:34
  • The naive way is to just compute the powers of 9 mod 17. If you find a 4 before a 1, you have your exponent. Otherwise, there's no solution. – Simon Goater Jan 06 '25 at 14:37
  • ah okay, than I did it for an exercise that requested me to find the order of 9 mod 17 and I found that the x I am looking for now is 6. How can I solve this problem in another way? – Irene Jan 06 '25 at 14:38
  • This operation is called discrete logarithm and there are many more complicated algorithms which work more efficiently for larger moduli. – Simon Goater Jan 06 '25 at 14:49
  • By using modular inverses like you have and arriving at the two equivalent congruences $9^{x+2} \equiv 1 \pmod{17}$ and $3^{2x+4} \equiv 1 \pmod{17}$, you can use Fermat's little theorem to get $x=6$ and $x=14$ quickly. I'm not positing this as an answer though because these definitely aren't the only solutions, but you can probably take this idea and use it to find the rest somehow. – mediocrevegetable1 Jan 06 '25 at 15:30
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Jan 06 '25 at 15:37
  • $9^x=4\pmod{17}$ is equivalent to $3^x=\pm 2\pmod{17}.$ I used WA to find the solutions $x=16n+a$ with $a\in {6,14}.$ It gives you at leat the hint to consider $x\pmod{16}$ if you want to solve it manually. – Marius S.L. Jan 06 '25 at 15:42

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\begin{align*} 9^{16n+a}&=4\pmod{17} \text{ with } 0<a<16\\ a=8m+b&\Longrightarrow 9^a= 9^b=4\pmod{17}\text{ with } 0<b<8 \end{align*} The only solution by checking all seven possibilities is $b=6$ and so $x=16n+8m+6$ or $x\in \{16n+6,16n+14\}.$

Marius S.L.
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