show this $\left(\left(\frac{p-1}{3}\right)!\right)^3+1\not\equiv 0\pmod p$ let $p\equiv 1\pmod 3$ is prime number,show that
$$\left(\left(\dfrac{p-1}{3}\right)!\right)^3+1\not\equiv 0\pmod p$$
I know this well
Show that $[(\frac{p-1}{2})!]^2+1\equiv0 \pmod p$
and I try use same idea to
$$-1\equiv (p-1)!=\left(\dfrac{p-1}{3}\right)!\cdot\left(\dfrac{p+2}{3}\cdot\dfrac{p+5}{3}\cdot\dfrac{p+p-2}{3}\right)\left(\cdot\dfrac{p+p+1}{3}\cdot\dfrac{p+p+4}{3}\cdots\dfrac{p+p+p-3}{3}\right)\equiv \left(\dfrac{p-1}{3}\right)!\cdot\left(\dfrac{2}{3}\cdot\dfrac{5}{3}\cdots\dfrac{p-2}{3}\right)\left(\dfrac{1}{3}\cdot\dfrac{4}{3}\cdots\dfrac{p-3}{3}\right)\pmod p$$
 A: Let $X$ be the elliptic  curve $u^3+v^3+w^3= 0$, which has the Weierstrass form $y^2 = x^3 - 432$. Now one can confirm the following facts (not necessarily easily).

*

*If $p \equiv 1 \bmod 3$, the number of points on $X$ has the form $1 + p - a_p$, where
$|a_p| < 2 \sqrt{p}$ (Weil bounds).


*If $p \equiv 1 \bmod 3$, then there is a congruence
$$a_p \equiv \frac{(p-1)!}{((p-1)/3)!^3} \bmod p,$$
Hence your requirement (given Wilson's theorem $(p-1)! \equiv -1 \bmod p$) is that $a_p \equiv 1 \bmod 3$, and thus $a_p = 1$ by the Hasse bounds. This comes by noting that one can count points modulo $p$ by evaluating the sum
$$\sum 1 - (u^3 + v^3 + z^3)^{p-1},$$
because this expression equals $1$ if $u^3+v^3+z^3=0$ and equals zero otherwise. Explicitly, if there are $(p+1-a_p)$ projective points, there are $1 + (p-1)(p+1-a_p)$ affine points in $[u,v,w]$, and this is $a_p \bmod p$. Hence
$$a_p \equiv - \sum (u^3 + v^3 + z^3)^{p-1} \equiv - \sum u^{3i} v^{3j} z^{3k} (p-1)!/(i! j! k!).$$
But the the power sums $\sum u^{3i} v^{3j} z^{3k}$ vanish unless $3i$, $3j$, and $3k$ are all positive and divisible by $p-1$, so the only term contributing a non-zero sum is the coefficient of $(uvz)^{p-1}$.


*There is a congruence $a_p \equiv 2 \bmod 3$ for primes $p \equiv 1 \bmod 3$. More on this below. (One proof: the elliptic curve $E$ has a rational $3$-torsion point, so $3$ divides $1+p-a_p$ and hence $a_p \equiv 2 \bmod 3$ when $p \equiv 1 \bmod 3$.)

These facts combined imply that the congruence never holds.
An additional fact is that any $p \equiv 1 \bmod 3$ can be written as
$$p = N(a + \omega b) = a^2  - a b +  b^2.$$
Here $\omega^3 = 1$ and $a + \omega b$ is unique up to multiplication by powers of $-1$ and $\omega$,
so there are six possible choices. But there is a unique choice which is congruent to $1 \bmod 3$ in the ring of Eisenstein integers, and for that choice, one has
$$a_p = \mathrm{Tr}(a + \omega b) = 2a - b,$$ and hence
$$\frac{1}{((p-1)/3)!^3} \equiv b - 2 a \bmod p.$$
This is a special case of the theory of complex multiplication, although known in these cases by Gauss via the relation to Jacobi sums.
More concretely, this implies that $3|b$ and $a \equiv 1 \bmod 3$, which certainly implies that $2a-b \equiv 2 \bmod 3$.
As an alternative (to see a related example where there are solutions), consider instead the congruence:
$$((p-1)/3)!^3 \equiv 1 \bmod p,$$
this is the same as asking that $a_p = -1$, or that $2a-b=-1$ and since $(a,b) \equiv (1,0) \bmod 3$, that $a = 1+3n$, or $p=7 + 27 n + 27 n^2$ for some integer $n$ (positive or negative). The first few primes with this property are
$$p = 7, 61, 331, 547, 195, 2347, \ldots $$
For example $61 = N(4 + 9 \omega)$ and $8 - 9 = -1$. But also $20! \equiv 47 \bmod 61$ and $47^3 \equiv 1 \bmod 61$.
These will be exactly the primes with $((p-1)/3)!^3 \equiv 1 \bmod p$. Presumably there are infinitely many such $p$.
