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 Showing results 1 to 14 of 14 Search took 0.01 seconds. Search: Posts Made By: Dr Sardonicus
 Forum: Miscellaneous Math 2022-03-23, 13:01 Replies: 26 Views: 2,335 Posted By Dr Sardonicus If n == 3 (mod 8) is prime, then... If n == 3 (mod 8) is prime, then Mod(Mod(1,n)+x,x^4 + 1)^n == Mod(Mod(1,n) + Mod(1,n)*x^3, x^4 + 1) coefficients of x and x^2 are 0 If n == 5 (mod 8) is prime, then Mod(Mod(1,n)+x,x^4 + 1)^n ==...
 Forum: Miscellaneous Math 2022-03-23, 00:07 Replies: 26 Views: 2,335 Posted By Dr Sardonicus By "your condition" I mean, any of Mod(Mod(1,n) +... By "your condition" I mean, any of Mod(Mod(1,n) + x, x^4 + 1)^n == Mod(1 + x^r, x^4 + 1) for n == r (mod 8), r =3, 5, or 7. Any of these conditions imply that Mod(2,n)^n = Mod(2,n). My...
 Forum: Miscellaneous Math 2022-03-22, 19:43 Replies: 26 Views: 2,335 Posted By Dr Sardonicus I did come up with a "reduced" version of part of... I did come up with a "reduced" version of part of your condition Mod(Mod(1,n)+x,1+x^4)^n = Mod(1, n)*(1 - x) for n == 5 (mod 8) First, when n == 5 (mod 8), lift(Mod(1 + x, 1 + x^4)^n) is a...
 Forum: Miscellaneous Math 2022-03-21, 13:14 Replies: 26 Views: 2,335 Posted By Dr Sardonicus For every integer r = 0 to 7, the sequence an =... For every integer r = 0 to 7, the sequence an = lift(Mod(x+1,x^4+1)^(8*n + r)), n = 0, 1, 2, ... satisfies the recurrence an+2+136*an+1 + 16*an = 0. Thus, for any given r, the coefficients of 1,...
 Forum: Miscellaneous Math 2022-03-20, 15:08 Replies: 26 Views: 2,335 Posted By Dr Sardonicus "That's the stupidest proof I've ever seen in my life!" I finally thought of a very simple, straightforward proof of the "zero-ness" of the coefficients of x^3, x^2, x and 1 in lift(Mod(x+1, x^4 + 1)^k) for k == 2, 0, 6 and 4 (mod 8) respectively. All...
 Forum: Miscellaneous Math 2022-03-18, 15:02 Replies: 26 Views: 2,335 Posted By Dr Sardonicus As you can see from the following, I have... As you can see from the following, I have obtained closed-form expressions for the coefficients of the degree-less-that-4 lift of Mod(1 + x, 1 + x^4)^n. They can be written as 4-term sums similar to...
 Forum: Miscellaneous Math 2022-03-17, 14:06 Replies: 26 Views: 2,335 Posted By Dr Sardonicus Letv=[Mod(1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2),... Letv=[Mod(1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2), Mod(-1/4*x^3 + 3/4*x^2 - 3/4*x + 1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2), Mod(-1/4*x^2 + 1/2*x - 1/4, x^4 - 4*x^3 + 6*x^2 - 4*x + 2), Mod(-1/4*x + 1/4, x^4...
 Forum: Miscellaneous Math 2022-03-17, 01:29 Replies: 26 Views: 2,335 Posted By Dr Sardonicus BTW, similar to your observation with... BTW, similar to your observation with Mod(Mod(1,n)*x + 1, x^4 + 1), Mod(Mod(1,n)*x + 2, x^2 + 1)^(n+1) == 5 implies 5^(n-1) == 1 (mod n) for n == 3 (mod 4) (assuming 5 does not divide n). That...
 Forum: Miscellaneous Math 2022-03-16, 12:17 Replies: 26 Views: 2,335 Posted By Dr Sardonicus Quite safe. Here's another way to look at this: ... Quite safe. Here's another way to look at this: If Mod(Mod(1,n)*x + 1, x^4 + 1)^n == Mod(1 - x, x^4 + 1) taking the norm gives (Mod(2, n))^n = 2. Well done, Sir!
 Forum: Miscellaneous Math 2022-03-16, 01:06 Replies: 26 Views: 2,335 Posted By Dr Sardonicus So you did. I just read "n" all the way through.... So you did. I just read "n" all the way through. Oops. No good ideas for this question, either. At least not yet. I'm pretty sure I know how to get closed formulas for the coefficients of...
 Forum: Miscellaneous Math 2022-03-15, 15:16 Replies: 26 Views: 2,335 Posted By Dr Sardonicus (checks, OK, it's equivalent to test proposed... (checks, OK, it's equivalent to test proposed earlier in thread). Evidently, none up to 10^8 since you've tested that far. I don't intend to do a numerical sweep above that point, so if you want...
 Forum: Miscellaneous Math 2022-03-15, 14:22 Replies: 26 Views: 2,335 Posted By Dr Sardonicus Hmm. Might be time to start thinking about how to... Hmm. Might be time to start thinking about how to construct composite n that "fool" the test. If you turn up pseudoprimes < 10^8, I'd say abandon that test. How many tests ...
 Forum: Miscellaneous Math 2022-03-14, 12:36 Replies: 26 Views: 2,335 Posted By Dr Sardonicus Just out of curiosity, up to what limit has this... Just out of curiosity, up to what limit has this been checked?
 Forum: Miscellaneous Math 2022-03-13, 18:51 Replies: 26 Views: 2,335 Posted By Dr Sardonicus I was able to prove that the indicated... I was able to prove that the indicated congruences indeed hold if n is a prime congruent to 3, 5, or 7 (mod 8). It made an interesting exercise. So there are no counterexamples to these with n...
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