## Re: sufficent proof for primes p:=x^2+x+1

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• A nice Sunday evening, ... generate. Perhaps you need some data more. I checked the primes p:=x^2+x+1 where x is a power of 2. x:=1; while TRUE do
Message 1 of 6 , Mar 12 12:14 PM
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A nice Sunday evening,

> Thank you Bernhard for that data, way beyond what I could
generate.

Perhaps you need some data more. I checked the primes p:=x^2+x+1
where x is a power of 2.

x:=1;
while TRUE do
p:=(x+1)*x+1;
if p mod 3 = 1 then
erg:=powermod (2, (p-1)/3, p);
res:=powermod (erg, 3, p);
if res=1 and (erg=x or erg=x^2) and isprime (p)=FALSE then
print (ifactor (x), p, ifactor (x+1));
end_if;
end_if;
x:=x*2;
end_while;

27 4
2 , 18014398643699713, 3 19 87211

81
2 , 5846006549323611672814741748716771307882079584257,

5
3 19 163 87211 135433 272010961

171
2 ,

895897896871121684222976912227377711248658198893860113275530965754434
39658\
67181184471134228436276477953

3 2
, 3 19 571 174763 160465489 19177458387940268116349766612211

243
2 ,

199791907220223502808422222706762643567910281130558153654986045416023
79129\
859977620592666523272986616017227171838989504031362210844729986994352
9473

6
, 3 19 163 1459 87211 135433 139483 272010961
10429407431911334611

all of the examples are where x+1 is a power of 3 with > 3

> And there is at least one interesting subplot: why is it that when
> p=3 the composites are generated when x+1 is a power of two (with
> just one known exception),

I found 3 other exception 2^81, 2^171, 2^243 besides.
The numbers get a little bigger. :-)

Nice Greetings from the primes
Bernhard
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