3 ms·
>pq mod r == ((p mod r) * (q mod r)) mod r that's just multiplication in Z_r...? edit: alright i guess technically it uses homomorphism from Z to Z_r but you
by anon_tor_12345 5y ago
>pq mod r == ((p mod r) * (q mod r)) mod r
that's just multiplication in Z_r...?
edit: alright i guess technically it uses homomorphism from Z to Z_r but you don't need to be able to write out the proof to intuit/believe it