Skip to main content
1 of 3
joro
  • 25.4k
  • 10
  • 66
  • 121

What is wrong with this counterexample to primality test assuming GRH?

From SMOOTH NUMBERS: COMPUTATIONAL NUMBER THEORY AND BEYOND Andrew Granville pp.13-14:

2j. Lenstra’s polynomial time test as to whether an integer that is conjecturally prime, is rigorously squarefree. (note that if the Generalized Riemann Hypothesis (GRH) holds and $a^{n-1} \equiv 1 \pmod{n}$ for all $a < 2 \log^2{n}$ ($*$) then $n$ is indeed prime).

Searching the web for about 20 minutes couldn't find reference for this.

Q1 What is a reference for this claim?

Alleged counter example:

Let $k=9981$ and $n=(6k+1)(12k+1)(18k+1)$.

$n$ has only three prime factors coming from the closed form form and according to Wikipedia it is Carmichael number. By the definition, the smallest $a$ s.t. $a^{n-1} \ne 1 \pmod{n}$ is $6k+1$, which is larger than the RHS of $*$.

Pari session:

? k=9981;n=(6*k+1)*(12*k+1)*(18*k+1)
%28 = 1288666276813009
? for(a=1,2*log(n)^2,b=Mod(a,n)^(n-1);if(b!=1,print(a)));
? \\nothing printed

Q2 What is wrong with the alleged counterexample?

joro
  • 25.4k
  • 10
  • 66
  • 121