How many keys are needed for symmetric encryption with $n$ participants, and why?

$\frac{n(n-1)}{2}$ keys are needed, because every pair of participants needs a unique shared secret.

* Pairwise symmetric keys grow quadratically; asymmetric key pairs grow linearly. *

For $n$ participants communicating pairwise:

• Each participant needs a key with every other participant
• Each participant has $n - 1$ keys
• But each key is shared by 2 people, so:

$$\text{Keys} = \frac{n \cdot (n-1)}{2}$$

Example: For $n = 100$ participants:

$$\frac{100 \times 99}{2} = 4{,}950 \text{ keys}$$

This is $O(n^2)$ growth, which becomes impractical at scale. This "key distribution problem" is one of the main motivations for asymmetric cryptography.

Note: If keys are directional ($K_{AB} \neq K_{BA}$), the count doubles to $n(n-1)$.

Go deeper:

• Key management — why quadratic key growth drives real-world key-distribution design.