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Search: Posts Made By: paulunderwood
Forum: Miscellaneous Math 2022-03-22, 22:14
Replies: 26
Views: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
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: 2,453
Posted By paulunderwood
Cool 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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