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It is known that each $p=3k+1$ has a unique representation of the form $$ p=\frac{L^2+27M^2}4, $$ up to sign change; see Wikipedia:.

The same Wikipedia page cites Gauss's theorem:

Under the same notation, we have $L\cdot k!^3\equiv 1\pmod p$.

[EDITED AND EXTENDED] As Geoff Robinson points out, this holds for an undetermined sign choice of $L$, so we need to work more.

Assume now that $k!\equiv -1\pmod p$; this yields that $L\equiv -1\pmod p$, so either $L=-1$ or $|L|\geq p-1$. The latter cannot hold, as $4p<(p-1)^2+27$; so $L=-1$.

Thus, $4p=1+27M^2$. Writing $M=2m-1$, we arrive at the expansion $$ p=\frac{1+3M\sqrt{-3}}2\cdot \frac{1-3M\sqrt{-3}}2 =\bigl((3m-1)+3\omega(2m-1)\bigr)\bigl((3m-1)+3\bar\omega(2m-1)\bigr) $$ in the Eisenstein ring $\mathbb Z[\omega]$.

The number $$ \pi=(3m-1)+3\omega(2m-1) $$ is a primary Eisenstein integer, hence by the cubic reciprocity law (see the same Wiki page) we have $$ \left(\frac2\pi\right)_3=\left(\frac \pi2\right)_3 =\left(\frac{m+1+\omega}2\right)_3=:\alpha, $$ We have $\alpha=\omega$ if $m$ is odd, and $\alpha=\bar\omega$ if $m$ is even.

If $m$ is odd, we have $$ r\equiv 2^k\equiv \omega \equiv\omega-(2\omega+1)\pi =\omega-(\omega-(9m-5))=9m-5\pmod \pi. $$ Similarly, $r\equiv 9m-5\pmod{\bar\pi}$, and therefore $r\equiv 9m-5\pmod p$. Hence, $r=9m-5$ (notice that $9m-5<p$), and $r\equiv 4\pmod 6$. This is excluded by the conditions of the question.

If $m$ is even, we similarly get $$ r\equiv \bar\omega \equiv\bar\omega+(2\omega+1)\pi =-1-\omega+(\omega-(9m-5)) =4-9m\pmod\pi $$ and hence $r\equiv 4-9m\pmod p$, so $r=p+4-9m\equiv 5\pmod 6$. This is also excluded.

P.S. Pitifully, I have no access to that book, but Wikipedia claims that Gauss's theorem is Exercise 7.9 in Lemmermeyer's book:

Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, ISBN 3-540-66957-4

It is known that each $p=3k+1$ has a unique representation of the form $$ p=\frac{L^2+27M^2}4, $$ up to sign change; see Wikipedia:.

The same Wikipedia page cites Gauss's theorem:

Under the same notation, we have $L\cdot k!^3\equiv 1\pmod p$.

[EDITED AND EXTENDED] As Geoff Robinson points out, this holds for an undetermined sign choice of $L$, so we need to work more; here it goes.

Assume now that $k!\equiv -1\pmod p$; this yields that $L\equiv -1\pmod p$, so either $L=-1$ or $|L|\geq p-1$. The latter cannot hold, as $4p<(p-1)^2+27$; so $L=-1$.

Thus, $4p=1+27M^2$. Writing $M=2m-1$, we arrive at the expansion $$ p=\frac{1+3M\sqrt{-3}}2\cdot \frac{1-3M\sqrt{-3}}2 =\bigl((3m-1)+3\omega(2m-1)\bigr)\bigl((3m-1)+3\bar\omega(2m-1)\bigr) $$ in the Eisenstein ring $\mathbb Z[\omega]$.

The number $$ \pi=(3m-1)+3\omega(2m-1) $$ is a primary Eisenstein integer, hence by the cubic reciprocity law (see the same Wiki page) we have $$ \left(\frac2\pi\right)_3=\left(\frac \pi2\right)_3 =\left(\frac{m+1+\omega}2\right)_3=:\alpha, $$ We have $\alpha=\omega$ if $m$ is odd, and $\alpha=\bar\omega$ if $m$ is even.

If $m$ is odd, we have $$ r\equiv 2^k\equiv \omega \equiv\omega-(2\omega+1)\pi =\omega-(\omega-(9m-5))=9m-5\pmod \pi. $$ Similarly, $r\equiv 9m-5\pmod{\bar\pi}$, and therefore $r\equiv 9m-5\pmod p$. Hence, $r=9m-5$ (notice that $9m-5<p$), and $r\equiv 4\pmod 6$. This is excluded by the conditions of the question.

If $m$ is even, we similarly get $$ r\equiv \bar\omega \equiv\bar\omega+(2\omega+1)\pi =-1-\omega+(\omega-(9m-5)) =4-9m\pmod\pi $$ and hence $r\equiv 4-9m\pmod p$, so $r=p+4-9m\equiv 5\pmod 6$. This is also excluded.

P.S. Pitifully, I have no access to that book, but Wikipedia claims that Gauss's theorem is Exercise 7.9 in Lemmermeyer's book:

Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, ISBN 3-540-66957-4

It is known that each $p=3k+1$ has a unique representation of the form $$ p=\frac{L^2+27M^2}4, $$ up to sign change; see Wikipedia:.

The same Wikipedia page cites Gauss's theorem:

Under the same notation, we have $L\cdot k!^3\equiv 1\pmod p$.

[EDITED] As Geoff Robinson points out, this holds for an undetermined sign choice of $L$, so we need to work more.

Assume now that $k!\equiv -1\pmod p$; this yields that $L\equiv -1\pmod p$, so either $L=-1$ or $|L|\geq p-1$. The latter cannot hold, as $4p<(p-1)^2+27$; so $L=-1$.

Thus, $4p=1+27M^2$. Writing $M=2m-1$, we arrive at the expansion $$ p=\frac{1+3M\sqrt{-3}}2\cdot \frac{1-3M\sqrt{-3}}2 =\bigl((3m-1)+3\omega(2m-1)\bigr)\bigl((3m-1)+3\bar\omega(2m-1)\bigr) $$ in the Eisenstein ring $\mathbb Z[\omega]$.

The number $$ \pi=(3m-1)+3\omega(2m-1) $$ is a primary Eisenstein integer, hence by the cubic reciprocity law (see the same Wiki page) we have $$ \left(\frac2\pi\right)_3=\left(\frac \pi2\right)_3 =\left(\frac{m+1+\omega}2\right)_3=:\alpha, $$ We have $\alpha=\omega$ if $m$ is odd, and $\alpha=\bar\omega$ if $m$ is even.

If $m$ is odd, we have $$ r\equiv 2^k\equiv \omega \equiv\omega-(2\omega+1)\pi =\omega-(\omega-(9m-5))=9m-5\pmod \pi. $$ Similarly, $r\equiv 9m-5\pmod{\bar\pi}$, and therefore $r\equiv 9m-5\pmod p$. Hence, $r=9m-5$ (notice that $9m-5<p$), and $r\equiv 4\pmod 6$. This is excluded by the conditions of the question.

If $m$ is even, we similarly get $$ r\equiv \bar\omega=-1-\omega \equiv-1-\omega+(2\omega+1)\pi =-1-\omega+(\omega-(9m-5)) =4-9m\pmod\pi $$ and hence $r\equiv 4-9m\pmod p$, so $r=p+4-9m\equiv 5\pmod 6$. This is also excluded.

P.S. Pitifully, I have no access to that book, but Wikipedia claims that Gauss's theorem is Exercise 7.9 in Lemmermeyer's book:

Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, ISBN 3-540-66957-4

It is known that each $p=3k+1$ has a unique representation of the form $$ p=\frac{L^2+27M^2}4, $$ up to sign change; see Wikipedia:.

The same Wikipedia page cites Gauss's theorem:

Under the same notation, we have $L\cdot k!^3\equiv 1\pmod p$.

[EDITED AND EXTENDED] As Geoff Robinson points out, this holds for an undetermined sign choice of $L$, so we need to work more.

Assume now that $k!\equiv -1\pmod p$; this yields that $L\equiv -1\pmod p$, so either $L=-1$ or $|L|\geq p-1$. The latter cannot hold, as $4p<(p-1)^2+27$; so $L=-1$.

Thus, $4p=1+27M^2$. Writing $M=2m-1$, we arrive at the expansion $$ p=\frac{1+3M\sqrt{-3}}2\cdot \frac{1-3M\sqrt{-3}}2 =\bigl((3m-1)+3\omega(2m-1)\bigr)\bigl((3m-1)+3\bar\omega(2m-1)\bigr) $$ in the Eisenstein ring $\mathbb Z[\omega]$.

The number $$ \pi=(3m-1)+3\omega(2m-1) $$ is a primary Eisenstein integer, hence by the cubic reciprocity law (see the same Wiki page) we have $$ \left(\frac2\pi\right)_3=\left(\frac \pi2\right)_3 =\left(\frac{m+1+\omega}2\right)_3=:\alpha, $$ We have $\alpha=\omega$ if $m$ is odd, and $\alpha=\bar\omega$ if $m$ is even.

If $m$ is odd, we have $$ r\equiv 2^k\equiv \omega \equiv\omega-(2\omega+1)\pi =\omega-(\omega-(9m-5))=9m-5\pmod \pi. $$ Similarly, $r\equiv 9m-5\pmod{\bar\pi}$, and therefore $r\equiv 9m-5\pmod p$. Hence, $r=9m-5$ (notice that $9m-5<p$), and $r\equiv 4\pmod 6$. This is excluded by the conditions of the question.

If $m$ is even, we similarly get $$ r\equiv \bar\omega \equiv\bar\omega+(2\omega+1)\pi =-1-\omega+(\omega-(9m-5)) =4-9m\pmod\pi $$ and hence $r\equiv 4-9m\pmod p$, so $r=p+4-9m\equiv 5\pmod 6$. This is also excluded.

P.S. Pitifully, I have no access to that book, but Wikipedia claims that Gauss's theorem is Exercise 7.9 in Lemmermeyer's book:

Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, ISBN 3-540-66957-4

It is known that each $p=3k+1$ has a unique representation of the form $$ p=\frac{L^2+27M^2}4, $$ up to sign change; see Wikipedia:.

The same Wikipedia page cites Gauss's theorem (where, I hope, $L$ is meant to be positive):

Under the same notation, we have $L\cdot k!^3\equiv 1\pmod p$.

Assume now that $k!\equiv -1\pmod p$; this yields that $L\equiv -1\pmod p$, so that $L\geq p-1$. But then $4p\geq (p-1)^2+27$, which cannot hold.

NB. The condition on the residue of $2^{k+1}$ seems to be useless.

It is known that each $p=3k+1$ has a unique representation of the form $$ p=\frac{L^2+27M^2}4, $$ up to sign change; see Wikipedia:.

The same Wikipedia page cites Gauss's theorem:

Under the same notation, we have $L\cdot k!^3\equiv 1\pmod p$.

[EDITED] As Geoff Robinson points out, this holds for an undetermined sign choice of $L$, so we need to work more.

Assume now that $k!\equiv -1\pmod p$; this yields that $L\equiv -1\pmod p$, so either $L=-1$ or $|L|\geq p-1$. The latter cannot hold, as $4p<(p-1)^2+27$; so $L=-1$.

Thus, $4p=1+27M^2$. Writing $M=2m-1$, we arrive at the expansion $$ p=\frac{1+3M\sqrt{-3}}2\cdot \frac{1-3M\sqrt{-3}}2 =\bigl((3m-1)+3\omega(2m-1)\bigr)\bigl((3m-1)+3\bar\omega(2m-1)\bigr) $$ in the Eisenstein ring $\mathbb Z[\omega]$.

The number $$ \pi=(3m-1)+3\omega(2m-1) $$ is a primary Eisenstein integer, hence by the cubic reciprocity law (see the same Wiki page) we have $$ \left(\frac2\pi\right)_3=\left(\frac \pi2\right)_3 =\left(\frac{m+1+\omega}2\right)_3=:\alpha, $$ We have $\alpha=\omega$ if $m$ is odd, and $\alpha=\bar\omega$ if $m$ is even.

If $m$ is odd, we have $$ r\equiv 2^k\equiv \omega \equiv\omega-(2\omega+1)\pi =\omega-(\omega-(9m-5))=9m-5\pmod \pi. $$ Similarly, $r\equiv 9m-5\pmod{\bar\pi}$, and therefore $r\equiv 9m-5\pmod p$. Hence, $r=9m-5$ (notice that $9m-5<p$), and $r\equiv 4\pmod 6$. This is excluded by the conditions of the question.

If $m$ is even, we similarly get $$ r\equiv \bar\omega=-1-\omega \equiv-1-\omega+(2\omega+1)\pi =-1-\omega+(\omega-(9m-5)) =4-9m\pmod\pi $$ and hence $r\equiv 4-9m\pmod p$, so $r=p+4-9m\equiv 5\pmod 6$. This is also excluded.

P.S. Pitifully, I have no access to that book, but Wikipedia claims that Gauss's theorem is Exercise 7.9 in Lemmermeyer's book:

Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, ISBN 3-540-66957-4