Vrftsjtryk

Let $p$ be a prime number. Is it true that there aren't any three different numbers $x,y,z$ such that $$x^3+x equiv y^3+y equiv z^3+z pmod p $$
with $x -y, y-z, z-x$, each of them cannot be divided by $p$ ?

If not, what are the conditions of $p$ so that the statement is true for prime number $p$ ?

I tried with $p=3,7$ and both of them are correct, so I think that $p equiv 3 pmod 4$ may satisfy the statement.

My other attempt: Assume by contradiction, there exist $x,y,z$ such that $$x^3+x equiv y^3+y equiv z^3+z pmod p$$
with $x -y, y-z, z-x$, each of them cannot be divided by $p$. Then $$x^2+xy+y^2 equiv y^2+yz+z^2 equiv z^2+zx+x^2 pmod p$$
thus $$x+y+z equiv 0 pmod p.$$

Here I am stuck. How can I solve this problem ?

(Sorry for my English)

Solutions exist for all primes $pge5,pneq7$.

As the OP observed, we have the Vieta relation $x+y+z=0$ as $x,y,z$ are the zeros of the cubic
$$
P(T)=T^3+T+c=(T-x)(T-y)(T-z)
$$
in the field $Bbb{F}_p$. Here $-c=-xyz$ is the shared value of $x^3+x,y^3+y$ and $z^3+z$ (treated as elements of $Bbb{F}_p$ turning congruences into equalities).

The relation $z=-x-y$ takes care of the quadratic term, and
we are well placed to take advantage of the degree of freedom to select $c$ any which way we wish. Let's concentrate on the linear term! Expanding $(T-x)(T-y)(T+x+y)$ tells us that
$$
(T-x)(T-y)(T+x+y)=T^3-T(x^2+xy+y^2)-xyz,
$$
so we want to be able to choose distinct elements $x,yinBbb{F}_p$ such that $x^2+xy+y^2=-1$.

This is possible whenever $p>3$.

Assume first that $pequiv1pmod3$. In this case there is a primitive cubic root of unity $omegainBbb{F}_p$. It satisfies the equation
$$
omega^2+omega+1=0.
$$
And that relation gives us the factorization
$$
a^2+ab+b^2=(a-omega b)(a-omega^2b).
$$
So we can select any two numbers $c,dinBbb{F}_p$ such that $cd=-1$. Then the linear system
$$
left{begin{array}{lcl}
a-omega b&=&c\
a-omega^2b&=&d
end{array}right.
$$
has a unique solution $(a,b)$. After all, its determinant is $omega-omega^2neq0$.

Then assume that $pequiv-1pmod3$. In this case $omega$ only exists in the extension field $Bbb{F}_{p^2}$. But, in that case we are dealing with the norm map
$$
N:Bbb{F}_{p^2}toBbb{F}_p, a-bomegamapsto (a-bomega)(a-bomega^2)=a^2+ab+b^2.
$$
By elementary properties of finite fields the norm is surjective, and takes each non-zero value in $Bbb{F}_p$ exactly $p+1$ times. In particular, there are $p+1$ pairs $(a,b)$ such that $a^2+ab+b^2=-1$.

The above argument did not concern the possibility that some of $x,y,z$ may be equal (i.e. $P(T)$ has a multiple root for the resulting $c$). If $x=y$, then $x^2+xy+y^2=3x^2$. If $-1/3$ is a quadratic residue, we need to rule out two possible values of $x$. If $x=-y-x$ then $y=-2x$, and again $3x^2=-1$. Finally, if $y=-y-x$ then $x=-2y$ we need to rule the solutions of $3y^2=-1$.
At most six pairs $(x,y)$ were ruled out. If $p>7$ then in the first case the number of pairs $(c,d)$ such that $cd=-1$ is high enough to leave some solutions. All the cases where we had repetitions among ${x,y,-x-y}$ lead to the presence of a square root of $-3inBbb{F}_p$, so the second case of $pequiv-1pmod 3$ is not affected.

The claim follows.

It may be worth noting that $p=7$ fails precisely because all the solutions of $a^2+ab+b^2=-1$, namely $(a,b)in{(1,3),(3,1),(3,3),(4,4),(4,6),(6,4)}$ lead to repetitions among ${a,b,-a-b}$. None of the six solutions of $cd=-1$ work!

well, if $p > 31$ and we can express
$$ p = u^2 + uv + 8 v^2 $$
with integers, then there are three distinct solutions to $t^3 + t equiv -1 pmod p$

     31,     47,     67,    131,    149,    173,    227,    283,    293,    349,

    379,    431,    521,    577,    607,    617,    653,    811,    839,    853,

    857,    919,    937,    971,   1031,   1063,   1117,   1187,   1213,   1237,

   1259,   1303,   1327,   1451,   1493,   1523,   1559,   1583,   1619,   1663,

   1721,   1723,   1741,   1879,   1931,   1973,   1993,   2003,   2017,   2153,

   2273,   2333,   2341,

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

? p = 47

%5 = 47

? factormod( x^3 + x + 1, p)

%6 = 

[Mod(1, 47)*x + Mod(12, 47) 1]



[Mod(1, 47)*x + Mod(13, 47) 1]



[Mod(1, 47)*x + Mod(22, 47) 1]



? p = 67

%7 = 67

? factormod( x^3 + x + 1, p)

%8 = 

[ Mod(1, 67)*x + Mod(4, 67) 1]



[ Mod(1, 67)*x + Mod(9, 67) 1]



[Mod(1, 67)*x + Mod(54, 67) 1]



? p=131

%9 = 131

? factormod( x^3 + x + 1, p)

%10 = 

[ Mod(1, 131)*x + Mod(56, 131) 1]



[ Mod(1, 131)*x + Mod(80, 131) 1]



[Mod(1, 131)*x + Mod(126, 131) 1]



? p=149

%11 = 149

? factormod( x^3 + x + 1, p)

%12 = 

[Mod(1, 149)*x + Mod(11, 149) 1]



[Mod(1, 149)*x + Mod(56, 149) 1]



[Mod(1, 149)*x + Mod(82, 149) 1]



?

well, I checked pretty high for primes $p$ such that, for some fixed $c = c(p),$ the relation $x^3 + x + c equiv 0 pmod p$ has three distinct roots $pmod p.$ As far as I can tell, this always happens unless $p = 3,7$

There are some patterns behind the scenes. When $p equiv 1 pmod 4$ we can use $c=0.$ When Legendre symbol $(p|7)=1$ we can use $c=2.$ When $p = u^2 + uv + 8 v^2$ we can use $c=1.$ When $p = u^2 + uv + 62 v^2$ or $p = 8u^2 + 3uv + 8 v^2$ we can use $c=3.$ When $p = 2u^2 + 2uv + 55 v^2$ we can use $c=4.$

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

3   WOW  

5  c:  0   roots:  0   2   3   

7   WOW  

11  c:  2   roots:  5   7   10   

13  c:  0   roots:  0   5   8   

17  c:  0   roots:  0   4   13   

19  c:  8   roots:  3   4   12   

23  c:  2   roots:  10   14   22   

29  c:  0   roots:  0   12   17   

31  c:  6   roots:  9   26   27   

37  c:  0   roots:  0   6   31   

41  c:  0   roots:  0   9   32   

43  c:  2   roots:  19   25   42   

47  c:  1   roots:  25   34   35   

53  c:  0   roots:  0   23   30   

59  c:  4   roots:  7   20   32   

61  c:  0   roots:  0   11   50   

67  c:  1   roots:  13   58   63   

71  c:  2   roots:  32   40   70   

73  c:  0   roots:  0   27   46   

79  c:  2   roots:  13   67   78   

83  c:  11   roots:  19   23   41   

89  c:  0   roots:  0   34   55   

97  c:  0   roots:  0   22   75   

101  c:  0   roots:  0   10   91   

103  c:  8   roots:  16   34   53   

107  c:  2   roots:  49   59   106   

109  c:  0   roots:  0   33   76   

113  c:  0   roots:  0   15   98   

127  c:  2   roots:  23   105   126   

131  c:  1   roots:  5   51   75   

137  c:  0   roots:  0   37   100   

139  c:  4   roots:  32   48   59   

149  c:  0   roots:  0   44   105   

151  c:  2   roots:  70   82   150   

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

This is not true. Consider $x=0$, $y=2$, and $z=3$ mod $5$ (this is a special case of saulspatz counterexample).