Friday, August 18, 2023, 19:30:01 + 6
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@ -72,3 +72,10 @@ We start by defining the document-class
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But you may be better off editing a document simply through [Overleaf](https://www.overleaf.com/project/6306feb35a99fd317a79ec3a) as well.
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![[Pasted image 20220825004820.png]]
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## Latex Cheatsheet:
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---
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- use `\` for spaces if not already handles
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- for the Obsidian plugin use `$$` to enable
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- for multi-integer exponentiation use `a^{34 + 65}`
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@ -0,0 +1,44 @@
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$$16^{54} \ mod \ 17=3^{24×54} \ mod \ 17 $$
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Why?
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[Public Key Cryptography: Diffie-Hellman Key Exchange (short version) - YouTube](https://www.youtube.com/watch?v=3QnD2c4Xovk) is a good video to understand asymmetric cryptography. There is a jump from 4:34 in the video that is not obvious to everyone.
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The jump is from 1654mod17 to 324×54mod17, but why? From the comments of video, I’m not the only one who is supprised by this jump.
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First let me introduce a formula:
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$$a^b\ mod \ p=((a \ mod \ p)^b) \ mod \ p \ \ \ \ (1)$$
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Then the proof:
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There must be one integer `n` to have
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$$a \ mod \ p=a−np \ \ \ \ (2)$$
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so
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$$((a \ mod \ p)^b) \ mod \ p=((a−np)^b) \ mod \ p$$
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With [Binomial theorem - Wikipedia](https://en.wikipedia.org/wiki/Binomial_theorem),
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$$(a−np)b=ab+(b1)ab−1(−np)+(b2)ab−2(−np)2+...+(−np)b$$
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We could see that all items are times of `p` except $a^b$,
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$$((a−np)bmodp)=(ab+(b1)ab−1(−np)+(b2)ab−2(−np)2+...+(−np)b)modp=abmodp+0+0+...+0$$
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Now we got
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$$((a−np)^b \ mod \ p)=a^b \ mod \ p$$
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Use formula (2)
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$$((amodp)b)modp=abmodp$$
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Now let a as $3^{24}$, and b as 54 in formula (1):
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$$3^{24×54}\ mod \ 17=((3^{24})^{54}) \ mod \ 17=((324mod17)54)mod17=1654mod17$$
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