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Here is the primality test I found for a number $n$ and the algorithm :

Step 1 : Let $n$ an odd number not divisible by $3$ or $5$ and greater than $5$.

Step 2 : Let $k$ an positive integer (here $k = 1$).

Step 3 : Let $a = (k-1)^2 + 3$.

Step 4 : Let $\left(\frac{n}{a}\right)$ denotes the Jacobi Symbol.

Step 5 : While $a = (k-1)^2 + 3$ is even or $\left(\frac{n}{a}\right)\neq1$ or $gcd(n,a+1) \neq 1$, add $1$ to $k$.

Step 6 : Let $B \equiv x^{n-1}$ mod $(n, f_a)$ where $f_a = x^2 - ax - a$

Step 7 : Let $T=x+n-a-1$ and $U=(n+1)/2 \cdot x + 2$

Step 8 : If $B \neq 1$ check if $B = T$ or $B = U$, if it is, return prime, if not, return composite

Step 9 : If $B = 1$ add $1$ to $k$ and redo step 3 to step 8.

Here is a PARI GP Code

k=1;
c(n,k) = {a=(k-1)^2+3;while(a%2!=1||kronecker(n,a)!=1||gcd(n,a+1)!=1,k++;a=(k-1)^2+3);
B=Mod(Mod(x,n),x^2-a*x-a)^(n-1);
T=x+n-a-1;U=((n+1)/2)*x + 2;
if(B!=1, if(B==T || B==U, print(n,",","probably prime"), 
print(n,",","composite"))); 
while(B==1, k++ && c(n,k))}


{for(n=1,10000,
if(n!=1&&n%2==1&&n%5!=0&&n%3!=0,c(n,k)))}

I have checked all primes and composites to $2^{31}$ and all pass the test.

I would like to know if this is only an coincidence or not. And if not, is there a way to prove it ?

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    $\begingroup$ Please make your PARI GP Code consistent with your steps. Why do you say $k=1$ in step 2 but assign $k=0$ in your PARI GP Code? Why do your introduce he variable $S=B$ in your PARI GP Code when there is no $S$ in your steps? Why is the variable $x$ in your steps and PARI GP Code never assigned a value? If the value of $x$ is unassigned, how can you assign a value to $\text{Mod}(x,n)$ for example? Have your also checked all primes to $10^9$? $\endgroup$ Commented Jun 3 at 15:53
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    $\begingroup$ Thanks. I'm not familiar with PARI GP Code. Is while(B==1, k++ && c(n,k)) the same thing as while(B==1, k++; c(n,k))? I'm used to using && as a logical and operation which doesn't seem to make sense to me in this statement. $\endgroup$ Commented Jun 3 at 17:47
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    $\begingroup$ Since you work modulo $n$, you can set $T=x-a-1$ and $U=x/2+2$. Also, $a$ is even when $k$ is even, so you can safely increase $k$ by 2 (instead of 1) to keep it odd. $\endgroup$ Commented Jun 3 at 18:14
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    $\begingroup$ In Step 5, what is the point of increasing $k$ when $\gcd(n,a+1)\ne1$? If $\gcd(n,a+1)\ne1$, you know $n$ is composite (assuming $a$ can’t get as high as $n$ itself?). $\endgroup$ Commented Jun 3 at 18:24
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    $\begingroup$ @Aurel-BG: Here is a cleaned version of your test, which for a given n returns 1 or 0 depending on its asserted primality: pastebin.com/DW9DHUGy $\endgroup$ Commented Jun 3 at 18:43

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As for many other similar tests, it's relatively easy to see why the proposed test cannot fail on prime $n$, but it may be hard to (dis)prove that there are no pseudoprimes (composites determined by the test as "probably prime").

In case of prime $n$ and a particular choice of $a$, this tests will get $B=1$, $B=U$ or $B=T$ depending on Jacobi symbol $\left(\frac{a^2+4a}n\right)$ being $1$, $0$, or $-1$, respectively.

As Martin Seysen pointed out, the case $\left(\frac{a^2+4a}n\right)=-1$, meaning that $f_a(x)$ is irreducible modulo $n$, corresponds to Quadratic Frobenius test, which in the given settings will get $x^{n+1}\equiv a\pmod{n, f_a(x)}$, implying that $x^{n-1}\equiv T\pmod{n, f_a(x)}$.

The case $\left(\frac{a^2+4a}n\right)=0$ means $a\equiv -4\pmod n$ and thus $f_a(x)\equiv (x+2)^2\pmod{n}$. Then $x^n = (-2+(x+2))^n \equiv (-2)^n \equiv -2\pmod{n, (x+2)^2}$, and thus $x^{n-1} \equiv U\pmod{n, (x+2)^2}$.

Finally, when $\left(\frac{a^2+4a}n\right)=1$, we have $f_a(x) = (x-r_1)(x-r_2)$ for some residues $r_1\ne r_2$ modulo $n$. Then $x^{n-1} \equiv r_i^{n-1} \equiv 1\pmod{n, x-r_i}$ for $i\in\{1,2\}$, and thus $x^{n-1}\equiv 1\pmod{n,f_a(x)}$.

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  • $\begingroup$ Thanks for you answer (and the pastebin) I noticed than the case $U = x/2 + 2$ seems only happens when the number $n$ is a prime of the form $x^2 + 7y^2$ by the way. $\endgroup$ Commented Jun 4 at 1:21
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    $\begingroup$ @Aurel-BG: If prime $n$ divides $a+4 = (k-1)^2 + 7$, it must be of the form $x^2+7y^2$. $\endgroup$ Commented Jun 4 at 2:16

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