[glaukopis]

The above argument might feel slightly wrong to people because it relies on the property that (g^a mod p) ^ b mod p == (g^ab mod p). The fact that this holds isn't too hard to figure out, but requires knowledge that pq mod r == ((p mod r) * (q mod r)) mod r, which again isn't too hard to figure out. However, at this point you're asking people to derive two basic number theory properties in a row, and notably "using modulus to write fizz buzz" (which is the level most non-math-major programmers are at wrt to modulus) is not the same thing as knowing how to do basic number theory, and proficiency in the former doesn't imply proficiency in the latter.

[anon_tor_12345]

>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