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 $?

Question

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)

Answer 1

Solutions exist for all primes $p\ge5,p\neq7$.

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 $x,y\in\Bbb{F}_p$ such that $x^2+xy+y^2=-1$.

This is possible whenever $p>3$.

Assume first that $p\equiv1\pmod3$. In this case there is a primitive cubic root of unity $\omega\in\Bbb{F}_p$. 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 $c,d\in\Bbb{F}_p$ 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 $\omega-\omega^2\neq0$.

Then assume that $p\equiv-1\pmod3$. 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 $$ N:\Bbb{F}_{p^2}\to\Bbb{F}_p, a-b\omega\mapsto (a-b\omega)(a-b\omega^2)=a^2+ab+b^2. $$ 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 $-3\in\Bbb{F}_p$, so the second case of $p\equiv-1\pmod 3$ is not affected.

The claim follows.

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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!

Answer 2

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

Answer 3

If $p=n^2+1$ (such as $5,17,37\dots$), then $x=0,\ y=n,\ z=(p-n)$ will solve your equivalence with all three terms congruent to $0$.

$x^3+x=0;\ y^3+y=n(n^2+1)=np\equiv 0 \mod{p}; z^3+z=(p-n)(p^2-2np+n^2+1)=(p-n)(p^2-2np+p)=p(p-n)(p-2n+1)\equiv 0\mod{p}$

$p$ cannot be of the form $n^2+1$ if $p\equiv 3\mod{4}$.

Answer 4

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]

?

Answer 5

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   

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