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garo · 2020-06-08, 11:43

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

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

2020-06-08, 11:43	#1
garo Aug 2002 Termonfeckin, IE 2768₁₀ Posts	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