20171201, 05:12  #1 
May 2004
2^{2}×79 Posts 
Pseudoprimes in special fields
Consider the special field: Mod(x^2+7).
Then ((15+7*sqrt(7))^1041)/105 is equal to an integer in this field i.e. 105 is a Fermat pseudoprime in the field under consideration. 
20171201, 06:12  #2 
Aug 2006
5,987 Posts 
I assume you're talking about the field \(\mathbb{Q}(\sqrt{7})\)? Or really, the ring of integers \(\mathcal{O}_{\mathbb{Q}(\sqrt{7})} = \left\{m + n\frac{1+\sqrt{7}}{2}\ \ m,n \in \mathbb{Z}\right\}\).
Yes, 105 is a pseudoprime to base 15+7*sqrt(7). There are 255 other bases in the ring \(\mathcal{O}_{\mathbb{Q}(\sqrt{7})}\) to which 105 is a pseudoprime, for example 20+7*sqrt(7) and 65/2+21*sqrt(7)/2. 
20171202, 17:46  #3 
Feb 2017
Nowhere
13534_{8} Posts 
I assume the ring in question is the ring R of integers in Q(sqrt(7)). As long as the modulus n is an odd integer greater than 1, 2 is invertible (mod n), and you can express residues in R/nR in the form Mod(Mod(a,n) + Mod(b,n)*x,x^2 + 7).
In the case n = 105 = 3*5*7, the invertible elements of (R/105R)* may be expressed as the direct product (R/3R)* x (R/5R)* x (R/7R)* which may easily be seen to be C_{8} x C_{24} x C_{42} The subgroup of elements of orders dividing 104 is then easily seen to be C_{8} x C_{8} x C_{2} which has order 128. 
20171203, 16:41  #4 
Feb 2017
Nowhere
2^{2}×5×13×23 Posts 
Again assuming we're working in the ring of integers R of Q(sqrt(7)), I submit for your amusement
n = 5632705 = 5*13*193*449 for which x^(n1) == 1 (mod n) for any x in R whose norm is relatively prime to n. There are smaller such composite n, but I chose this one because of the factors 5 and 13 :D 
20171205, 03:26  #5  
May 2004
2^{2}×79 Posts 
Quote:
Last fiddled with by devarajkandadai on 20171205 at 03:30 Reason: want to be more explicit 

20171205, 03:52  #6 
"Dana Jacobsen"
Feb 2011
Bangkok, TH
3^{2}·101 Posts 
Made me think of https://arxiv.org/pdf/1307.7920.pdf and https://arxiv.org/pdf/1706.01265.pdf.
They may or may not actually have any connection to what you're discussing. 
20171205, 14:00  #7  
Feb 2017
Nowhere
175C_{16} Posts 
On 20171201, 05:12,
Quote:
I don't care whether this is utter incompetence or deliberate trollery. I'm tired of it. 

20171206, 01:46  #8 
May 2004
100111100_{2} Posts 
Charles, you have shown me how to identify rational integer bases for pseudoprimality of 105. Each of these can be split to form algebraic integer bases for the sameexample: Have already shown that 22 can be split as 15+7*I, 15 + 7*I*sqrt(7) and many with shape 15 +7*sqrt(y) where y is prime excepting 3 and 5.Next 8 can be split as 15 7*I or 157*sqrt(y) where y is prime. Have tested a few. Similarly 29 can be split as 15 + 14*sqrt(y).

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