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Old 2020-06-08, 11:43   #1
garo
 
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Aug 2002
Termonfeckin, IE

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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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