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);}
[/CODE]

:surprised:

Morning after... A quick 1 second scan of Jan Feitsma's 2-PSP list yields the fraud tst(619033*670619).