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Prove that there exist a prime $p\equiv 1(\text{mod }3)$ such that $x^3\equiv 2(\text{mod }p)$ for some $x\in\mathbb{Z}$.

I'm trying to use the fact that $a$ is a cube modulo $p$ if and only if $a^{\frac{p-1}{3}}\equiv 1(\text{mod }p)$, but I can't prove that $2^{\frac{p-1}{3}}\equiv 1(\text{mod }p)$ is necessarily true for some prime $p$. Any hints?

  • 4
    $4^3=64\equiv 2\pmod {31}$ – 2017-02-12
  • 1
    Do you actually want to prove the existence of ONE prime with this property ? Or do you want to prove the property for every prime of the form $3k+1$ ? – 2017-02-12
  • 1
    It's for one prime. For $p=7=3(2)+1$, we have that $2$ is not a cube modulo $7$ – 2017-02-12

1 Answers 1

3

I guess the best way to investigate is just to factor $x^3 - 2$ for integers $x> 1.$ See when prime factors are $1 \pmod 3.$ In this little output I can see $31, 43, 439, 499$

    2      6 = 2 * 3
    3     25 = 5^2
    4     62 = 2 * 31
    5    123 = 3 * 41
    6    214 = 2 * 107
    7    341 = 11 * 31
    8    510 = 2 * 3 * 5 * 17
    9    727 = 727
   10    998 = 2 * 499
   11   1329 = 3 * 443
   12   1726 = 2 * 863
   13   2195 = 5 * 439
   14   2742 = 2 * 3 * 457
   15   3373 = 3373
   16   4094 = 2 * 23 * 89
   17   4911 = 3 * 1637
   18   5830 = 2 * 5 * 11 * 53
   19   6857 = 6857
   20   7998 = 2 * 3 * 31 * 43
   21   9259 = 47 * 197

Take a prime $$ p = u^2 + 27 v^2 $$ in integers $u,v.$ For example, $31.$ Just calculate the cubes of $1,2,3,4,5,6,7 \pmod{31},$ see what happens.

      1,     31,     43,    109,    127,    157,    223,    229,    277,    283,
    307,    397,    433,    439,    457,    499,    601,    643,    691,    727,
    733,    739,    811,    919,    997,   1021,   1051,   1069,   1093,   1327,
   1399,   1423,   1459,   1471,   1579,   1597,   1627,   1657,   1699,   1723,
   1753,   1777,   1789,   1801,   1831,   1933,   1999,   2017,   2089,   2113,
   2143,   2179,   2203,   2251,   2281,   2287,   2341,   2347,   2383,   2671,
   2689,   2731,   2749,   2767,   2791,   2833,   2917,   2953,   2971,

=====================================================================

The primes $q \equiv 1 \pmod 3$ that do not work are those $q = 4 u^2 + 2 u v + 7 v^2,$

 7,     13,     19,     37,     61,     67,     73,     79,     97,    103,
139,    151,    163,    181,    193,    199,    211,    241,    271,    313,
331,    337,    349,    367,    373,    379,    409,    421,    463,    487,
523,    541,    547,    571,    577,    607,    613,    619,    631,    661,
673,    709,    751,    757,    769,    787,    823,    829,    853,    859,
877,    883,    907,    937,    967,    991,   1009,   1033,   1039,   1063,

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