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Old 2020-06-08, 11:43   #1
garo
 
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Default Modular arithmetic query

Apologies if this is very basic. Could anyone tell me why

(g^{a}\ mod\ p) \cdot (g^{b}\ mod\ p) \ mod\ p\equiv g^{(a+b)\ mod\ (p-1)}\ mod\ p

Last fiddled with by garo on 2020-06-08 at 11:47
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Old 2020-06-08, 11:53   #2
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Because: https://en.wikipedia.org/wiki/Euler%27s_theorem
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Old 2020-06-08, 12:07   #3
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Right. I got that far but couldn't make the connection. How do I get from the totient function to (a+b) mod (p-1)?
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Old 2020-06-08, 12:09   #4
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Quote:
Originally Posted by garo View Post
Right. I got that far but couldn't make the connection. How do I get from the totient function to (a+b) mod (p-1)?
For primes the totient function is simply p-1. So multiples of p-1 in the exponent can be ignored.
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Old 2020-06-08, 12:15   #5
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Gotcha. Thanks for your help. Not sure why I was making it more complicated in my head.

Last fiddled with by garo on 2020-06-08 at 12:16
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