3

I have a discrete log that I need to solve to aid in a Cryptography problem, that deals with both programming and mathematics, so I was unsure where to post this problem, feel free to move me if needed. I must use Index-Calculus to solve this discrete log. Here's the setup, my attempt, and my code (producing the wrong 'r' value).

Let $p=10007$. $5$ is a primitive root. It can be shown that $L_5(2)=6578, L_5(3)=6190, L_5(7)=1301$. Use these facts to find $L_5(100)$

So, I have the fact that

$100\times5^r\equiv7\times3\times2\mod 10007$

$\implies L_5(100)\equiv-r+L_5(7)+L_5(3)+L_5(2)\mod 10006$

And $L_5(7)+L_5(3)+L_5(2)$ are known, so I now just need to find a $r$ s.t.

$100\times5^r\equiv7\times3\times2\equiv42\mod 10007$

So I wrote a program that basically just loops through $j$ for $0\leqslant j\lt p$, and then calculates $100*5^j\equiv ans \mod 10007$ and checks if $ans = 42$. When it is equal to 42, it prints the j used. This is producing an answer of $r=j=378$, however this is NOT correct, as $100\times5^{378}\not\equiv 42\mod 10007$. The correct answer is $r=j=911$.

Where have I gone wrong? Or else, is there an easier way to solve this problem? (Using Index-Calculus, I can solve it using Baby Step Giant Step, or Pohlig-Hellman algorithm.

Bill Dubuque
  • 282,220
justanotheruser
  • 97
  • It seems you aren't clear about how discrete logs work. If my answer doesn't help clarify that then let me know and I can elaborate. – Bill Dubuque Mar 27 '19 at 23:14

2 Answers2

4

It is called a discrete log because $\,L(xy)\equiv L(x)+L(y)\ \pmod{p\!-\!1},\ $ therefore

$$\begin{align} L(2)&\,\equiv\, 6579,\ L(5)\equiv 1\\[.2em] \Rightarrow\, L((2\cdot 5)^2) &\,\equiv\, 2(L(2)+L(5)) \\[.2em] &\equiv\, 2(6578+1)\end{align}\qquad \qquad\qquad\quad\!$$

Remark $ $ While you can ignore the logs and instead work with powers of $5$ as in Robert's answer, this likely is completely opposite of the goal of the exercise, which is likely intended to teach you how to convert such multiplicative problems into simpler additive problems in $\,\Bbb Z_{p-1} = $ integers $\bmod p\!-\!1\ $ (which is the raison d'etre of this index calculus).

Bill Dubuque
  • 282,220
  • 2
    Whether you write it as $L(2^2 \cdot 5^2) = 2 L(2) + 2 $ or $2^2 \cdot 5^2 = 5^{2 \cdot L(2) + 2}$ is a matter of taste. The actual computation is the same. – Robert Israel Mar 27 '19 at 22:53
  • @Robert Of course. But I don't agree that is is simply a matter of "taste". Rather. it is a matter of pedagogy. That the student titled it "discrete log solve" surely means they were taught to solve these problems using discrete logs. Your answer is of little help in that regard since it completely eliminates the log viewpoint. – Bill Dubuque Mar 27 '19 at 23:10
2

$100 = 2^2 \cdot 5^2$. If $2 \equiv 5^{6578} \bmod p$, then $2^2 \cdot 5^2 \equiv 5^{2 \cdot 6578 + 2} \equiv 5^{3152}\bmod p$.

Robert Israel
  • 470,583