Primality Formula Conjecture
To test any $(6x-1)$ numbers for primality.
$$ 4^{3x-1} \bmod (6x-1) $$
If that is equal to 1 then $(6x-1)$ is prime.
To test any $(6x+1)$ numbers for primality.
$$ 4^{3x} \bmod (6x+1) $$
If that is equal to 1 then $6n+1$ is prime.
I am unsure if this holds true for infinity. I have only tested around 200 numbers by hand. It quickly becomes difficult because of the exponent, and I am also not a professional Mathematician, I just like to play with numbers. So please be patient with me.
Backstory
I stumbled upon this while fooling around with numbers and the Collatz conjecture. I was first studying
$$ \frac{3x+1}{2} $$
After which I became interested in finding what numbers of $x\bmod 6 = 5$ or $x\bmod 6 = 1$ became a power of 2 within the Collatz Conjecture. So I shifted things around a bit and create this formula.
$$ y = \frac{2^{x}-1}{3} $$
That resulted in this series.
| X | Y |
|---|---|
| 1 | $\frac{1}{3}$ |
| 2 | $1$ |
| 3 | $2\frac{1}{3}$ |
| 4 | $5$ |
| ... | ... |
This was too noisy with fractions every other result, so I removed the noise by using 4.
$$ y = \frac{4^{x}-1}{3} $$
| X | Y |
|---|---|
| 1 | $1$ |
| 2 | $5$ |
| 3 | $21$ |
| 4 | $85$ |
| ... | ... |
I then became interested in knowing the factors of those numbers.
| X | Y | Prime Factors |
|---|---|---|
| 1 | $1$ | 1 |
| 2 | $5$ | 5 |
| 3 | $21$ | 3, 7 |
| 4 | $85$ | 5, 17 |
| 5 | $341$ | 11, 31 |
| 6 | $1365$ | 3, 5, 7, 13 |
| 7 | $5461$ | 43, 127 |
| 8 | $21845$ | 5, 17, 257 |
| 9 | $87381$ | 3, 7, 19, 73 |
| 10 | $349525$ | 5, 11, 31, 41 |
| 11 | $1398101$ | 23, 89, 683 |
| ... | ... | ... |
I then noticed a pattern and created this formula.
$$ \frac{4^{3x-1}-1}{3} $$
| X | Y | Prime Factors |
|---|---|---|
| 1 | $5$ | 5 |
| 2 | $341$ | 11, 31 |
| 3 | $21845$ | 5, 17, 257 |
| 4 | $1398101$ | 23, 89, 683 |
| 5 | $89478485$ | 5, 29, 43, 113, 127 |
| 6 | $22906492245$ | 1, 3, 5, 7, 13, 19, 37, 73, 109 |
| ... | ... | ... |
I noticed when $\frac{4^{3x-1}-1}{3}$ was divisible by $(6x-1)$ then it was a prime number. I then continued testing this conjecture with hundreds of primes. It seemed to work.
I don't know how to go about proving this. I'm not practiced in proofs being just a math hobbyist. It would be amazing to hear your feedback about this conjecture.
Thanks for reading.
If $p=6x-1$ is prime,
then by Fermat's little theorem $4^{3x-1}=2^{6x-2}=2^{(6x-1)-1}\equiv 1\pmod{6x-1}$.
But the converse does not hold for a pseudoprime base $2$ such as $341=6\times57-1$:
$4^{3\times57-1}=2^{(6\times57-1)-1}=2^{340}\equiv1\pmod{341=6\times57-1}$,
though $341=11\times31$ is not prime.