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I encountered a number theory problem when doing my research:

1.I want to know whether or not there are infinitely many primes $p$ satistying $gcd(\frac{p-1}{6},6)=1$, such that $6$ is a cubic residue mod $p$, but $2$ and $3$ are not cubic residues mod $p$? If there are, can we give a expression of $p$?

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    $\begingroup$ For (2) the answer is surely yes --- the first few such primes occur for k=0,-2,3,-4,4,5,-6,-9,10,11 --- but (as with every other case of Bunyakowsky's conjecture in degree 2 or greater) no technique is known that could prove it. Fortunately this is not needed for a positive answer to (1), which will follow by applying Chebotarev's density theorem to a suitable extension of the third cyclotomic field. $\endgroup$ Commented Mar 26, 2019 at 1:27
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    $\begingroup$ Please, Zuo, look up Bunyakowsky's Conjecture. You will find there is no one-variable quadratic for which it has been proved that the quadratic takes on infinitely many prime values. $\endgroup$ Commented Mar 26, 2019 at 8:55
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    $\begingroup$ Doesn't a positive answer for (1) follow immediately from the Cebotarev density theorem? $\endgroup$ Commented Mar 26, 2019 at 20:27
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    $\begingroup$ Posted to m.se, math.stackexchange.com/questions/3163220/… $\endgroup$ Commented Apr 2, 2019 at 2:43
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    $\begingroup$ After the latest edit, the body of the question doesn't match the title. And it's still not clear whether you have looked up Buyakowsky yet. $\endgroup$ Commented Apr 16, 2019 at 12:59

1 Answer 1

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You want primes $$ p = 7 x^2 + 6 xy + 36 y^2 $$ with integers $x,y.$ Notice that such primes are represented by $4 s^2 + 2 st + 7 t^2$ and $7 u^2 + 3 uv + 9 v^2,$ so both $2$ and $3$ are not cubic residues. Here one could just write $s = 3y, t = x,$ for the other $u=x, v = 2y.$

Meanwhile, in $ p = 7 x^2 + 6 xy + 36 y^2 $ we could take $x=1$ and $y = 6k$ to arrive at your $7 + 36 k + 1296 k^2$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
7 6 36
Discriminant  -972

Modulus for arithmetic progressions? 
3
Maximum number represented? 
3000

      7,     37,    139,    163,    181,    241,    313,    337,    349,    379,
    409,    421,    541,    571,    607,    631,    751,    859,    877,    937,
   1033,   1087,   1123,   1171,   1291,   1297,   1447,   1453,   1483,   1693,
   1741,   1747,   2011,   2161,   2239,   2311,   2371,   2473,   2539,   2647,
   2677,   2707,   2719,   2857,
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