i think it is vaguely plausible that there might be two primes that have perfect complementarity in their roots but i doubt it could be found in the range of values we can calculate in any sane amount of time. so yeah, all products of prime roots would almost certainly be irrational as well.
Discussion
Can one find p1 where n is sufficiently small and p1 and p2 are large primes when p2 and mod(sqrt(p1*p2),n) are known?
honestly i have no idea about primes except for how they are the basis of cryptographic encodings, exploiting their properties in combination with clockwork arithmetic.
so that equation sounds a lot like reversing a signature to find the secret key, but i'm not sure.
I stated it wrong. Here it is corrected:
Can one find p1 if p1 and p2 are large primes when p2 and a finite subset of mod(sqrt(p1*p2),1) are known?
I know this isn't exactly a trapdoor function, but it seems like it could be converted to one
i don't think it's a trapboor function but rather a form of asymmetric cipher, it's not too distantly different from an elliptic curve.
https://wikimedia.org/api/rest_v1/media/math/render/svg/3dbe6cab1bc2c7f1c99757dc6e5d7a517cf9b4f8
y^2=x^3+a*x+b
your equation is:
p1^2 = (p1*p2)%x)
mod 1 is um... 1. i'm using the simple modulo notation % meaning "remainder", which is sorta like one step of a modulo, more or less the same operation, at least, one cycle of a clockwork arithmetic system. that mod value tells you the potential number of values that could be valid for y, relative to how p2 relates to it.
i'm used to working with these things in code not so much in math. the finite field is a different shape to the field of real numbers.
Mod 1 is intended to drop the whole number part and only yield the irrational fractional component
i'm pretty sure anything mod one is the whole number part and to remove it you then need to subtract this from the result, so you'd have to repeat the formula with a subtract
p1^2 = (p1*p2)-((p1*p2) mod 1)
i forget what the precedence of mod is, i think it's commutative to multiply/divide thus the braces. also, mod(dividend, divisor) is programming function notation, i'm pretty sure in math notation you use mod as an infix.
i'm pretty foggy on all these things to be honest, i rarely have to decipher math notation to write algorithms, usually someone else already did that and i just copy/adapt it.
or have i got that backwards. lol! i'm alwaays getting things backwards and off-by-one.