Diffie Hellman Problem

Question

I know that in Diffie Hellman, the final key (from Bob's point of view the final key is calculated as follows)

K_B = (g^x mod n)^y mod n, where

x represents Alice's private no.
y represents Bob's private no.
g and n the two public nos.

which can be evaluated as

K = (g^xy mod n) OR (g^yx mod n), where

x represents Alice's private no.
y represents Bob's private no.
g and n the two public nos.

My question is that How does K_B = (g^x mod n)^y mod n evaluates to K = (g^xy mod n) OR (g^yx mod n). Does the mod operator has a property where (g^x mod n)^y mod n evaluates to g^xy mod n?

Ok so (g^x mod n).(g^x mod n)...(g^x mod n)...(y times) basically means (g^(y.x) mod n) and that is a property right? — Zaid Khan, Jul 10 '16 at 07:48
I'm really sorry but I missed a mod n in (g^x mod n)^y (I have now edited the question) so should your answer be (a mod n).(b mod n) = (ab mod n) mod n? — Zaid Khan, Jul 10 '16 at 09:59

score 0 · Answer 1 · answered Jul 10 '16 at 06:51

I think you have some misconceptions about what $\text{mod}$ means. Let $p$ be an integer, then the "object" $x\text{ mod }p$ is the equivalence class of $x$ in $\mathbb{Z}/p\mathbb{Z}$, that is to say the image of $x$ under the quotient map $$\mathbb{Z}\longrightarrow\mathbb{Z}/p\mathbb{Z},$$ which for simplicity we'll write $[x]$. It is easy to check that indeed $[x]^a = [x^a]$, which is the property you wanted.

Remark: all numbers considered here are integers.

@Bateman What? No, I meant exactly what I wrote. – Daniel Robert-Nicoud Jul 10 '16 at 22:09 — Daniel Robert-Nicoud, Jul 10 '16 at 22:09

Diffie Hellman Problem

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