2021-11-16, 23:54 | #1 |
Sep 2002
Database er0rr
3973_{10} Posts |
1+1 selfridges test for 1/8 of numbers
I was playing around while I waiting for some results to come in...
For n=3 mod 4 such that kronecker(-3,n)==-1 and kronecker(-7,n)==-1 test Mod(2,n)^((n-1)/2)==kronecker(2,n) and Mod(-7,n)^((n-1)/2)==-1. That's it! It is 1+1 selfridges. I have tested it for n < 7*10^9 so far. Pretty good, eh? Can you fool it? For convenience here is the test: Code:
{tst(n)=n%4==3&&kronecker(-3,n)==-1&&kronecker(-7,n)==-1&& Mod(-7,n)^((n-1)/2)==-1&&Mod(2,n)^((n-1)/2)==kronecker(2,n);} Morning after... A quick 1 second scan of Jan Feitsma's 2-PSP list yields the fraud tst(619033*670619). Last fiddled with by paulunderwood on 2021-11-17 at 08:13 |
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