The Math

Public Parameters

Alice and Bob first agree on two public numbers:

• p: a large prime number
• g: a generator (a number with special properties)

Everyone can know these. They’re not secret.

Let’s Use Real Numbers

Alice’s Side

Step 1: Alice picks a secret number.

$a = 6$

This is her private key. She never shares it.

Step 2: Alice computes her public value.

$A = g^a \mod p$

$\phantom{A} = 5^6 \mod 23$

$\phantom{A} = 15625 \mod 23$

$\phantom{A} = 8$

She sends A = 8 to Bob. Everyone can see this.

Bob’s Side

Step 1: Bob picks a secret number.

$b = 15$

This is his private key. He never shares it.

Step 2: Bob computes his public value.

$B = g^b \mod p$

$\phantom{B} = 5^{15} \mod 23$

$\phantom{B} = 19$

He sends B = 19 to Alice. Everyone can see this.

Creating the Shared Secret

Now comes the magic.

Alice computes:

$K = B^a \mod p$

$\phantom{K} = 19^6 \mod 23$

$\phantom{K} = 2$

Bob computes:

$K = A^b \mod p$

$\phantom{K} = 8^{15} \mod 23$

$\phantom{K} = 2$

Same answer! Both get K = 2.

This is their shared secret. They can now use it as a key for symmetric encryption.

Why Do They Get the Same Answer?

Alice computes: $B^a = (g^b)^a = g^{ba}$

Bob computes: $A^b = (g^a)^b = g^{ab}$

Since $ab = ba$, they get the same result.

What Eve (the Attacker) Sees

Eve is watching the whole conversation. She knows:

Why Can’t Eve Calculate K?

To find K, Eve would need either $a$ or $b$.

She knows $A = 8$ and that $A = 5^a \mod 23$.

So she needs to find $a$ where $5^a \mod 23 = 8$.

This is the Discrete Logarithm Problem.

For small numbers like 23, she could just try all possibilities. But for a 600-digit prime? There’s no known efficient algorithm.

Diffie-Hellman’s security rests on the difficulty of the discrete logarithm problem.