Notepad/enter/Proof of a formula for modulo.md

$$16^{54} \ mod \ 17=3^{24×54} \ mod \ 17 $$

Why?

[Public Key Cryptography: Diffie-Hellman Key Exchange (short version) - YouTube](https://www.youtube.com/watch?v=3QnD2c4Xovk) is a good video to understand asymmetric cryptography. There is a jump from 4:34 in the video that is not obvious to everyone.

The jump is from 1654mod17 to 324×54mod17, but why? From the comments of video, I’m not the only one who is supprised by this jump.

First let me introduce a formula:

$$a^b\ mod \ p=((a \ mod \ p)^b) \ mod \ p \ \ \ \ (1)$$

Then the proof:

There must be one integer `n` to have

$$a \ mod \ p=a−np \ \ \  \  (2)$$

so

$$((a \ mod \ p)^b) \ mod \ p=((a−np)^b) \ mod \ p$$

With [Binomial theorem - Wikipedia](https://en.wikipedia.org/wiki/Binomial_theorem),

$$(a−np)b=ab+(b1)ab−1(−np)+(b2)ab−2(−np)2+...+(−np)b$$

We could see that all items are times of `p` except $a^b$,

$$((a−np)bmodp)=(ab+(b1)ab−1(−np)+(b2)ab−2(−np)2+...+(−np)b)modp=abmodp+0+0+...+0$$

Now we got

$$((a−np)^b \ mod \ p)=a^b \ mod \ p$$

Use formula (2)

$$((amodp)b)modp=abmodp$$

Now let a as $3^{24}$, and b as 54 in formula (1):

$$
 \\ 

\\begin{align}
 3^{24×54}\ mod \ 17=((3^{24})^{54}) \ mod \ 17 \\ \\&=  ((3^{24} \ mod \17)^{54})mod 17 


\end{align}
= 16^{54}mod17$$