Quadratic Frobenius test

The quadratic Frobenius test (QFT) is a probabilistic primality test to determine whether a number is a probable prime. It is named after Ferdinand Georg Frobenius. The test uses the concepts of quadratic polynomials and the Frobenius automorphism. It should not be confused with the more general Frobenius test using a quadratic polynomial – the QFT restricts the polynomials allowed based on the input, and also has other conditions that must be met. A composite passing this test is a Frobenius pseudoprime, but the converse is not necessarily true.

Concept

Grantham's stated goal when developing the algorithm was to provide a test that primes would always pass and composites would pass with a probability of less than 1/7710.Template:Rp

The test was later extended by Damgård and Frandsen to a test called extended quadratic Frobenius test (EQFT).

Algorithm

Let Template:Mvar be a positive integer such that Template:Mvar is odd, and let b and c be integers such that $\left(\frac{b^2+4c}{n}\right)=-1$ and $\left(\frac{-c}{n}\right)=1$, where $\left(\frac{\cdot }{\cdot }\right)$ denotes the Jacobi symbol. Set $B=50000$. Then a QFT on Template:Mvar with parameters (Template:Mvar, Template:Mvar) works as follows:

(1) Test whether one of the primes less than or equal to the lower of the two values $B$ and $\sqrt{n}$ divides Template:Mvar. If yes, then stop: Template:Mvar is composite.

(2) Test whether $\sqrt{n}\in \mathbb{Z}$. If yes, then stop: Template:Mvar is composite.

(3) Compute $x^{\frac{n+1}{2}}mod(n,x^2-bx-c)$. If $x^{\frac{n+1}{2}}\notin \mathbb{Z}/n\mathbb{Z}$, then stop: Template:Mvar is composite.

(4) Compute $x^{n+1}mod(n,x^2-bx-c)$. If $x^{n+1}\equiv ̸-c$, then stop: Template:Mvar is composite.

(5) Let $n^2-1=2^{r}s$ with Template:Mvar odd. If $x^s\equiv ̸1mod(n,x^2-bx-c)$, and $x^{2^{j}s}\equiv ̸-1mod(n,x^2-bx-c)$ for all $0\le j\le r-2$, then stop: Template:Mvar is composite.

If the QFT does not stop in steps (1)–(5), then Template:Mvar is a probable prime.

(The notation $A\equiv Bmod(n,f(x))$ means that $A-B=H(x)\cdot n+K(x)\cdot f(x)$, where H and K are polynomials.)

See also

• Integers modulo n
• Multiplicative group of integers modulo n