20200213, 01:37  #1 
"Sam"
Nov 2016
311 Posts 
Agrawal's conjecture revisited
While the AKS test was published almost decades ago, I was wondering if Agrawal's conjecture might be true if some variant or other condition is added to it?
Here is what I have figured out so far, but I'm stuck on a few proofs. Does anyone got any further ideas? Last fiddled with by carpetpool on 20200213 at 01:41 Reason: Changed link 
20200213, 05:08  #2  
"Sam"
Nov 2016
311 Posts 
Quote:
Try this link. 

20200213, 15:17  #3 
Feb 2017
Nowhere
2×7×13×19 Posts 
Remarks on Agrawal's Conjecture indicates the conjecture is likely false.

20200213, 22:33  #4  
"Sam"
Nov 2016
467_{8} Posts 
Quote:
Assuming the falsehood of the original conjecture, what I was trying to illustrate is that counterexamples should have very specific properties, namely that the only counterexamples (with r prime) should be Carmichael numbers of order(m) where m = (r1)/2. There is no proof (that I know of), but there is evidence that this should be true. There should also exist upper bounds on either the Carmichael numbers of order(m) < x (and show that it is 0 for some x), or there is a lower bound such that the smallest Carmichael numbers of order(m) > L. If these two arguments are true, then there should exist a modified version of the conjecture that is true as we will have conditions for which n must be prime. Last fiddled with by carpetpool on 20200213 at 22:33 

20200213, 23:18  #5 
"Sam"
Nov 2016
311 Posts 
[1]
n is a primitive root mod r, and Agrawal's conjecture holds for (n, r), then n is either prime, or a Carmichael number of order (r1)/2. [2] There exist a lower bound such that the smallest Carmichael number of order(m) > L_{m}. If these two arguments are true, then there should exist a modified version of the conjecture that is true as we will have necessary and sufficient conditions for which n must be prime. Assuming the truth of these two arguments, and more specifically, the assumption that the lower bound L (as defined above), exists: [3] Find a prime r which n is a primitive root (mod r). Then check that (x  1)^n = x^n  1 mod (n, x^r  1), n < L_{m}, and (r  1) > 2*m. If so, then n is (or should be) prime. The problem with this test (apart from the unproven assumptions) is that r may be significantly large (for instance, if L = log^2(n) by the GRH). This would make this test (if true) impractical for large numbers. In particular, the complexity is dependent on the size of r. As pointed out earlier, for any individual prime q, n will be a prime or Carmichael number of order (q1)/2 if n is a primitive root mod q (condition [1]). Now for q_{2}, check that the conditions outlined in [1] are true. If so, n must also be (prime) or Carmichael number of order (q_{2}1)/2. Assume n is not prime. This implies that n is a Carmichael number of order lcm((q1)/2, (q_{2}1)/2) which is equivalent to λ(r)/4 where λ(r) is the Carmichael function of r. We can r by the conditions outlined in [3] as long as λ(r) > 4*m. However, this would be more practical as we will only have two to run [3] with two smaller primes q and q_{2} versus r, and this could decrease the complexity (and running time) drastically. What about r's with more factors? It is feasible to find a set of small primes [q, q_{2}, q_{3},... q_{i}] for which n is a primitive root mod every prime in the set. Then r is their product and as long as λ(r) > 2^i*m, tests in [3] can be applied to n. m is defined in [2] to be the smallest integer such that there is no Carmichael number of order m ≤ n. While I had a hard time attempting to put this all together it appears to be worthwhile to investigate the truth of [1] and [2] if not [3], at the very least. If you have any remarks, or (want) clarification, please let me know. Last fiddled with by carpetpool on 20200213 at 23:19 
20200215, 20:06  #6  
"Sam"
Nov 2016
311 Posts 
Quote:
(x  1)^(t*n + n) = x^n  1 mod(n, x^r  1) where t = 2*r*(n^((r  1)/2)  1) if jacobi(n  r) = 1 or (x  1)^(t*n + n) = x^n  1 mod(n, x^r  1) where t = (n^((r  1)/2)  1) if jacobi(n  r) = 1 if n^2 ≠ 1 mod r and r > 5. As there are infinitely many values of w such that (x  1)^w = x^n  1 mod(n, x^r  1), proving that w = t*n + n is the smallest such integer (apart from n itself) is false: n = 61, r = 7 So now I seem to have dug a deeper hole into solving this problem. We need a lower bound B such that (x  1)^w ≠ x^n  1 mod(n, x^r  1) if w is any integer (n < w < B). 

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