I am trying to calculate Alice and Bob's shared key by hand without the use of a calculator as I feel this is an important trait when progressing into cryptography.

I understand you can use the square and multiply method however we are being taught a shortcut method which I don't quite understand fully.

Question Example:

Alice and Bob use the DH protocol with p = 19,g = 2 and secrets a = 6 and b = 8. What key do they agree on?

They have given us this process without much explanation: $$ K = 2^{6×8} ≡8^{2×8} ≡7^8 ≡11^4 ≡7^2 ≡11 \pmod{19} $$

$2^{6×8}$, not sure how to put multiplication as power for the above problem.

If someone could explain in-depth the shortcut process completed above, step by step. I would really appreciate it

I understand some parts, like $g^a*b \pmod{19} = 6 = 2^3 = 8$, however I get slightly confused from there.

  • $\begingroup$ I polished some of the question, but left the last paragraph as is. $\endgroup$
    – fgrieu
    Nov 10 '21 at 9:32
  • $\begingroup$ Thankyou so much, In regards to your handheld calculations, i was wondering, where you did 2^48-18*2 - Why did you multiply it 2 at the end, and is that 2 the Generator? So is it g^48 mod(19-1)*g And Another thing, how did you know to use 19 and 215. This is where i struggle with the most. $\endgroup$ Nov 10 '21 at 9:43
  • $\begingroup$ In my $48-18\times2=12$, the $2$ is $\lfloor48/18\rfloor$, unrelated to $g$. This is to compute $48\bmod18$, and that modulus is $p-1$. In my $4096-19\times215=11$ I use $19$ because that's the modulus $p$. My $215$ was obtained (digit by digit) as$\lfloor4096/19\rfloor$. This is to compute $4096\bmod19$. I've added some of this in my answer. $\endgroup$
    – fgrieu
    Nov 10 '21 at 14:38
  • $\begingroup$ @fgrieu unfortunately this is a crossposted question with math. anthonymicheals1, you should need to learn that this is is not a good ethic in so $\endgroup$
    – kelalaka
    Nov 10 '21 at 18:19
  • $\begingroup$ If the answer is good you can upvote (min rep 15 required, that you have passed), if the answer satisfies you can accept it. This is the way of Stack Overflow. $\endgroup$
    – kelalaka
    Nov 11 '21 at 18:03

Recall that for $\forall x\in\mathbb N$, $\forall m,u,v\in\mathbb N^*$, it holds ${(x^u\bmod m)}^n\equiv{(x^u)}^v\equiv x^{u\times v}\pmod m$, where $y\equiv x\pmod m$ means $m$ divides $x-y$, and $x\bmod m$ is the uniquely defined integer $y$ such that $0\le y<m$ and $y\equiv x\pmod m$.

The shared secret is $K=(g^a\bmod p)^b\bmod p=(g^b\bmod p)^a\bmod p$, or equivalently $K=g^{a\times b}\bmod p$. We are tasked to evaluate this for $a=6$, $b=8$, $g=2$, $p=19$.

The method in the question goes: $$\begin{array}{} K&={(2^6\bmod19)}^8\bmod19&&=2^{6\times8}\bmod19\\ &=2^{(3\times2)\times8}\bmod19&={(2^3)}^{2\times8}\bmod19&=8^{2\times8}\bmod19\\ &={(8^2)}^8\bmod19&=64^8\bmod19\\ &&=(64-19\times3)^8\bmod19&=7^8\bmod19\\ &={(7^2)}^4\bmod19&=49^4\bmod19\\ &&=(49-19\times2)^4\bmod19&=11^4\bmod19\\ &={(11^2)}^2\bmod19&=121^2\bmod19\\ &&=(121-19\times6)^2\bmod19&=7^2\bmod19\\ &=49\bmod19&=49-19\times2&=11\\ \end{array}$$ and that (keeping the rightmost column) can be condensed to: $$K\equiv2^{6\times8}\equiv8^{2\times8}\equiv7^8\equiv11^4\equiv7^2\equiv11\pmod{19}\ \text{ thus }\ K=11$$

If I was to compute this without calculator, I would write this as $K=2^{48}\bmod19$, then use Fermat's Little Theorem. It says that when $p$ is prime and $g$ is not a multiple of $p$, if holds $g^{p-1}\bmod p=1$. That allows to reduce modulo $(p-1)$ any exponent of $g$ when computing modulo $p$. The complete calculation goes: $$\begin{array}{} K&=2^{6\times8}\bmod19&&=2^{48}\bmod19\\ &=2^{48\bmod(19-1)}\bmod19&=2^{48-18\times2}\bmod19&=2^{12}\bmod19\\ &=4096\bmod19&=4096-19\times215&=11\end{array}$$

Note: In $48-18\times2=12$, the $2$ is obtained as the quotient $\lfloor48/18\rfloor$, much like in $4096-19\times215=11$ the $215$ is $\lfloor4096/19\rfloor$.

Actual cryptography uses integers much too large for reliable human computation; e.g. $p$ could be 2048-bit, that is 617 decimal digits.


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