- A beautifull day,

I have examine the prime numbers which lay on the polynom

f(x)=x^2+x+1

My hypothese was :

if 2^[(p-1)/3] = x or 2^[(p-1)/3] =x^2,

then p is a prime .

The hypothes is not sufficent, because i found some counterexamples.

But these counterexamples are all of the same form.

They all have the form that x+1 is a potence of 2.

[MuPad]

x:=2;

while TRUE do

p:=x^2+x+1;

if p mod 3 = 1 then

erg:=powermod (2, (p-1)/3, p);

if (erg=x or erg=x^2) and isprime (p)=FALSE then

print (p, x+1, ifactor (x+1), erg, x);

end_if;

end_if;

x:=x+1;

end_while;

6

4033, 64, 2 , 63, 63

8

65281, 256, 2 , 255, 255

12

16773121, 4096, 2 , 4095, 4095

16

4294901761, 65536, 2 , 65535, 65535

18

68719214593, 262144, 2 , 262143, 262143

20

1099510579201, 1048576, 2 , 1048575, 1048575

I am looking for a proof that the counterexamples only appears,

if x+1 is like 2^n

Some thougths could be helpfull:

p:=x^2+x+1=x*(x+1)+1

Remember i check out if 2^[(p-1)/3] is similar to x or x^2

There are three possibilities:

1. The (x+1)-root of 1 is 1 <=> 2^(p-1)/(x+1)=1 mod p

2. The x-root of 1 is 1 <=> 2^(p-1)/x=1 mod p

3. There is a f divisor of x and g divisor of x+1 with

f*g>squareroot of p-1 so that 2^(p-1)/(f*g) = 1

So far so good, but how can i proof that the 3. case only appears if

x+1 is a power of 2.

Would be nice from you if you have a clever idea and share it to me

Nice greetings from the primes

Bernhard - 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

I found 3 other exception 2^81, 2^171, 2^243 besides.

> p=3 the composites are generated when x+1 is a power of two (with

> just one known exception),

The numbers get a little bigger. :-)

Nice Greetings from the primes

Bernhard