Mike had difficulty posting and asked me to post this for him:

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> Is the above, in essence, your conjectural method?

Yes.

Here is my function of 3Nov10:-

{

is_eligible(n)=

local(f,ind,p,res=1);

f=factor(n);

for(ind=1,#f[,1],

if(f[ind,2]>1,res=0;break(1)); \\not squarefree

p=f[ind,1];

if( ((n-1)%(p-1))<>0,

if( !((((n-1)%(p+1))==0)&&(((n+1)%(p-1))==0)),res=0;break(1));

); \\end if

); \\end for ind

res;

}

Now, how to prove this is kosher?...

Mike

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I have already posted the "sufficient" part of the proof.

The "necessary" part is awaited ...

David (pp Mike)