Fermat prime number test without Fermat pseudoprime numbers?

THE PRIME NUMBER SERIES 25

Assumption: There are functions that do not produce Fermatian pseudoprimes. This means that all numbers of such a function with a passed Fermat prime number test do not have the result ‘possibly prime’, but the result is ‘prime’ when a Fermat prime number test is passed.

The Fermat prime number test is performed with:

$a^{p-1} \equiv 1 \mod{p}$

So be it $k \in \mathbb{N}$ and $k \ge 0$ .

The following functions $p_{pot}(k)$ are provided by both prime numbers and compound numbers and, in Fermat’s prime number test, Fermat pseudoprimes:

$2k+3$

$4k+3$

$4k+5$

$6k+5$

$6k+7$

To avoid Fermat pseudoprimes, no composite number of a function to be defined must pass the Fermat prime number test.

An example of a function that presumably does not produce Fermatian pseudoprimes is

$p_{pot} = 4k^2 + 12k + 7$

I tested from $k=0$ to $k=10^9$ with the base a=2 and did not find a Fermat pseudoprime. In the ranges from $k=0$ to $k=10^7$ I tested with bases 3, 4, 5 and 7 and also did not find a Fermat pseudoprime number.

Furthermore, I randomly tested values up to k=10^10000 without Fermat’s pseudoprimes occurring.

Another example is Euler’s function

$p_{pot} = k^2 + k + 41$

I tested from $k=0$ to $k=10^9$ with the base a=2 and did not find a Fermat pseudoprime.

Munich, 16 May 2024

Gottfried Färberböck

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Fermat prime number test without Fermat pseudoprime numbers?