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 Showing results 1 to 13 of 13 Search took 0.01 seconds. Search: Posts Made By: paulunderwood
 Forum: Miscellaneous Math 2022-03-22, 22:14 Replies: 26 Views: 2,335 Posted By paulunderwood I don't see why it has to be base 2 Fermat PRP;... I don't see why it has to be base 2 Fermat PRP; The norm is -1.
 Forum: Miscellaneous Math 2022-03-22, 19:36 Replies: 26 Views: 2,335 Posted By paulunderwood Transforming the test of x over x^2-136*x+16, I... Transforming the test of x over x^2-136*x+16, I have now checked Feitsma's 2^64 PSP-2 list against: (n%8==3||n%8==5)&&Mod(Mod(x,n),x^2-1154*x+1)^((n+1)/2)==1. Given the test...
 Forum: Miscellaneous Math 2022-03-20, 21:13 Replies: 26 Views: 2,335 Posted By paulunderwood Powers of x over x^2-136*x+16 are interesting. ... Powers of x over x^2-136*x+16 are interesting. Mod(Mod(x,n),x^2-136*x+16)^(n+1)==16 is good for n%8==3 and n%8==5 I can also check Mod(Mod(x,n),x^2-136*x+16)^((n+1)/4)==2 for n%8==3, and...
 Forum: Miscellaneous Math 2022-03-17, 13:22 Replies: 26 Views: 2,335 Posted By paulunderwood Computing 5-PSPs up to 2^64 would be a mammoth... Computing 5-PSPs up to 2^64 would be a mammoth task involving many dedicated years of running efficient code on big hardware. I think getting the list of 2-PSPs involved some mathematical tricks that...
 Forum: Miscellaneous Math 2022-03-16, 18:07 Replies: 26 Views: 2,335 Posted By paulunderwood FWIW, not a peep out of: ... FWIW, not a peep out of: {forstep(n=5,1000000,8, if(!ispseudoprime(n), for(a=1,(n-1)/2, if(gcd(a,n)==1, Det=Mod(a,n)^4+1; ...
 Forum: Miscellaneous Math 2022-03-16, 05:45 Replies: 26 Views: 2,335 Posted By paulunderwood The companion matrix of x^4+1 is: ?... The companion matrix of x^4+1 is: ? cm=([0,0,0,-1;1,0,0,0;0,1,0,0;0,0,1,0]) [0 0 0 -1] [1 0 0 0] ...
 Forum: Miscellaneous Math 2022-03-15, 23:29 Replies: 26 Views: 2,335 Posted By paulunderwood n%8==5 tested up to n<10^10. I will take it to... n%8==5 tested up to n<10^10. I will take it to the next exponent.
 Forum: Miscellaneous Math 2022-03-15, 15:23 Replies: 26 Views: 2,335 Posted By paulunderwood I did say "any k". Powers k never seem to map x+1... I did say "any k". Powers k never seem to map x+1 to 1-x. HTH.
 Forum: Miscellaneous Math 2022-03-15, 15:05 Replies: 26 Views: 2,335 Posted By paulunderwood Considering n%8==5 with... Considering n%8==5 with Mod(Mod(x+1,n),x^4+1)^n==1-x as the main test, can you find any k such that Mod(Mod(x+1,n),x^4+1)^k==1-x for composite n?
 Forum: Miscellaneous Math 2022-03-15, 14:51 Replies: 26 Views: 2,335 Posted By paulunderwood That is easy for testing x+1 over x^4+1; It is... That is easy for testing x+1 over x^4+1; It is 8/2=4 times the number of Selfridges of x over x^2-a*x+1 (with (a^2-4 | n)==-1). I have not computed the function for x^2^k+1, but it gets big!
 Forum: Miscellaneous Math 2022-03-14, 13:17 Replies: 26 Views: 2,335 Posted By paulunderwood I have checked the n%3==4 test over x^2+1 up to... I have checked the n%3==4 test over x^2+1 up to 2^50. The others I tend to test up to 10^8 -- takes a minute or two. Furthermore, for n%4==3 I can test base x+2 over x^2+1, for n%8==5 base x+1...
 Forum: Miscellaneous Math 2022-03-14, 04:47 Replies: 26 Views: 2,335 Posted By paulunderwood The test over x^2+1 can be computed with 2... The test over x^2+1 can be computed with 2 Selfridges. Over x^4+1 it is 8 Selfridges by combining the difference of squares: ? Mod(s*x^3+t*x^2+u*x+v,x^4+1)^2 Mod((2*v*s...
 Forum: Miscellaneous Math 2022-03-13, 13:39 Replies: 26 Views: 2,335 Posted By paulunderwood mod x^4+1 You may be aware that no one has found a counterexample to the test Mod(Mod(x+2,n),x^2+1)^(n+1)==5 for n%4==3. I propose the test Mod(Mod(x+1,n),x^4+1)^(n-1) for odd n with the following results ...
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