
#1
Mar312, 03:22 PM

P: 891

As Dodo noted, my prior test excluded some primes of the form 8n +/ 3, e.g. 29. I discovered a new property of the recursive series used in the prior test to allow a correction that apparently includes all primes of the form 8n +/ 3 but apparently includes no composites. The property of the recursive series (defined by S_0 = 2, S_1 = 3, S_n = 6S_(n1)  S_(n2)  6) is that;
[tex]\sum_{i=0}^{2n} \binom{2n}{i}S_{kn+i} + 3*(2^{n}1)*2^{2n1} = 2^{3n} * S_{k}[/tex] Now if 2n = P+1 and P is prime of the form 8n +/ 3 and k = (P1)/2, the side on the right = 0 mod P and only the terms [tex]S_{1}, (P+1)*S_{0},(P+1)*S_{P1} \{and} S_{p}[/tex] not divisible by P on the left are . My experience is that S(1) = 3, S(0) = 2, S(P1) = P*S + 10 and S(P) = P*R + 3. Thus my new conjecture is that If and only If 6*2^((P+1)/2) equals 12 mod P, where P is a number of the form 8n +/ 3, then P is prime. I checked all 45 of my false hits, F, under 3 million, i.e. where S_((F1)/2) = 0 mod F. However, I hadn't yet gone back and checked whether any composite meets this latter test and also does not have a zero at S_((F1)/2). Edit I found ten false hits under 500,000. All were composites F such that S((F1)/2) > 0 mod F. Thus there is a need to further check that in the series S_0 = 2, S_1 = 3, S_n = 6*S_(n1)  S_(n2)  6 that S((P1)/2) = 0 mod P to determine that P is prime. None the less, both of these checks take far less time than checking that each and every term S_i is not equal 0 for i < (p1)/2 plus you have the advantage that all primes of the form 8n +/ 3 apparently meet the test. 



#2
Mar512, 10:14 AM

P: 891

So the prime test for P of the form 8n +/ 3 is in two parts:
A) If P is prime then 2^((P1)/2) = 1 mod P B) If P is prime then S((P1)/2)** = 0 mod P ** S is the recursive series S_0 = 2, S_1 = 3, S_n = 6*S_(n1)  S_(n2)  6 I did some more exploring and came up with a similar test for primes of the form 8n +/ 1 A) If P is prime then 2^((P1)/2) = 1 mod P B) If P is prime then R((P1)/2)** = 0 mod P ** R is the recursive series R_0 = 0, R_1 = 1, R_n = 6*R_(n1)  R_(n2) + 2, which is OEIS A001108. Both discoveries came from the following observations. . I created the R' type series by taking R'_i as the sum of two adjacent S terms + 1 and dividing by 2, i.e. R'_i = (S_i + S_(i+1) + 1)/2 which is the series R'_0 = 3, R'_1 = 7, having the recurrence R'_n = 6*R'_(n1)  R'_(n2)  4. Then I noted that S_(i+1) equals (R_i + R_(i+1) + 2)/4 for all i. From this I derived the summation property that I mentioned in the first post. Then I decide to check the R' series mod primes of the form 8n +/ 1 and discovered that R'_((P1)/2) = 3 mod P. So I subtracted 3 from each term and got each term = 0 mod 4. Therefore, I divided each term by 4. What is interesting is that each term of the R series taken as an argument of a triangular number yields the square triangular numbers, that is R*(R+1)/2 is always a perfect square. With the above discoveries a proof of these tests might be possible. Have the tests A already been proven? I would like to verify those frist if possible. 



#3
Mar512, 02:18 PM

P: 688

Hi, ramsey,
This follows from Euler's criterion (in that page, take a=2). And 2 is a 'quadratic residue' mod p (meaning, there is another number whose square is congruent to 2 mod p) whenever the prime is of the form 8ką1 (this statement is called the second suplement to the law of quadratic reciprocity). 


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