4 ms·
No. This proof says nothing about the security of RSA. I don't know anyone seriously trying to break RSA by factoring polynomials. It's pretty easy to test a bu
by wbhart 8y ago
No. This proof says nothing about the security of RSA. I don't know anyone seriously trying to break RSA by factoring polynomials. It's pretty easy to test a bunch of random polynomials and pretty quickly find that this just isn't going to work. You don't need a proof to know this empirically, which is why it isn't a serious attack.
Even if most polynomials were irreducible, which this work doesn't show, that wouldn't rule out having a method that cleverly, but efficiently writes the numbers as polynomials that aren't irreducible.
Also, according to the article, the result is only for polynomials with 0 or 1 coefficients, which aren't the only polynomials, obviously!