-1

I am fairly new to number systems and notations. I have a problem where I need to find $c$ when it is an exponent:

$a ≡ b^{c} \mod 43$ given that $0 \leq c < 43$ and, $a ≡ 27 \mod 43$ and $b ≡ 19 \mod 43$. ($a$ and $b$ are integers).

I know that the procedure is to solve the right-hand side, by replacing the values:

$ 27 ≡ 19^{c} \mod(43) $

But then, $c$ is on the right-hand side. Any approach arithmetic approach to take in this case?

is it allowed to do: $27^{-c} ≡ 19 \mod 43$ ???

Bill Dubuque
  • 282,220
user102859
  • 139
  • 2
    No, $27^{-c}$ isn't even guaranteed to be an integer (nor is $27^{1/c}$ which is what you would actually need to write). I think in common parlance this problem is called the discrete logarithm. – Ian Sep 29 '20 at 15:55

2 Answers2

0

$27^{-c}$ is not necessarily an integer as mentioned by Ian in the comments.

Generally you would use trial and error (for values of $c$) or quadratic reciprocity.

First by Fermat's little theorem we have that $$19^{42}\equiv 1 \pmod{43} $$

Now compute some powers of $19$ modulo $43$

$$19^2\equiv 17$$ $$19^3 \equiv 22$$ $$19^4 \equiv 31$$ $$\cdot\cdot\cdot$$ $$19^9 \equiv 27$$ $$19^{10} \equiv 40$$ $$\cdot\cdot\cdot$$

We see that $$27\equiv 19^{9}=19^{c}\pmod {43}$$

Thus we have $$c\equiv 9 \pmod{42}.$$

Alessio K
  • 10,679
-1

As Ian mentions in a comment, this is the Discrete Logarithm Problem, and it is hard. Its difficulty is one of the reasons that the RSA cryptosystem is secure.

For numbers this small, the best way to solve it is simply to enumerate the powers of $19$ until you hit $27$ (or $-27$). In this case, it doesn't take you too long.

TonyK
  • 68,059