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The given linear congruence is to be solved:

$$\gcd:(3;29)=1 |17$$ $$3x+29y=1 \iff 3x \equiv 1 \pmod{29} \iff y \equiv 3^{-1} \pmod{29}$$

With the extended Euclidean algorithm one obtains: \begin{align*} 29 & = 3 \cdot 9+2\\ 3 & = 2 \cdot 1+1\\ 2 & = 2 \cdot 1 \end{align*}

By means of backward resolution: \begin{align*} 1 & = 3-2 \cdot 1\\ & = -29+10 \cdot 3\\ & = -29+10(3-29)\\ & = 10 \cdot 3-1 \cdot 29 \end{align*}

$-1 \equiv 3^{-1} \pmod{29}$

$$\implies 3x \equiv 17 \pmod{27} \iff x \equiv -1 \cdot 3x \pmod{29} \equiv -1 \cdot 17 \pmod{29}$$

Are my considerations correct so far and how do I arrive at the solution $x$?

N. F. Taussig
  • 76,571
  • This is hard to follow. What does "$gcd:(3;29)=1|17$" mean? And $29$ is not equal to either $3\times9 +23$ nor $2\times 1$. In any case, isn't the inverse of $3\pmod {29}$ obviously $10$? – lulu May 26 '23 at 10:32
  • gcd(3,29)=1|17 mean that the result 1 divides the number 17, which is why the task can be solved at all. –  May 26 '23 at 10:36
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    You meant to write $\implies 3x \equiv 17 \pmod{29}$ . – N. F. Taussig May 26 '23 at 10:38
  • Writing run on equations is always confusing. $(3,29)$ is not equal to $1|17$ in any sense so don't write an equality between them. In any case, as I mentioned, the inverse of $3\pmod {29}$ is $10$ by inspection, no Euclidean algorithm needed. But your application of that algorithm does confirm $10$ as the inverse, not $-1$. – lulu May 26 '23 at 10:40
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    This MathJax tutorial explains how to typeset mathematics on this site. – N. F. Taussig May 26 '23 at 10:55
  • The easiest solution is to recognize that $17\equiv -12 \bmod 29$ immediately yielding $x\equiv -4 \equiv 25 \bmod 29$ – Keith Backman May 26 '23 at 14:47
  • See the linked dupes for most all known methods. – Bill Dubuque May 27 '23 at 01:22
  • 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 May 27 '23 at 01:35

1 Answers1

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You correctly applied the extended Euclidean algorithm to show that $1 = 10 \cdot 3 - 1 \cdot 29$. However, this does not imply that $-1$ is the multiplicative inverse of $3$ modulo $29$. Observe that $$1 \equiv 10 \cdot 3 - 1 \cdot 29 \equiv 10 \cdot 3 \pmod{29}$$ since $1$ and $10 \cdot 3$ differ by a multiple of $29$. Hence, $3^{-1} \equiv 10 \pmod{29}$.

Therefore, \begin{align*} 3x & \equiv 17 \pmod{29}\\ 10 \cdot 3x & \equiv 10 \cdot 17 \pmod{29}\\ x & \equiv 170 \pmod{29}\\ x & \equiv 5 \cdot 29 + 25 \pmod{29}\\ x & \equiv 25 \pmod{29} \end{align*}

Check: If $x \equiv 25 \pmod{29}$, then $3x \equiv 3 \cdot 25 \equiv 75 \equiv 2 \cdot 29 + 17 \equiv 17 \pmod{29}$.

N. F. Taussig
  • 76,571