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In his lovely answer at Positive primes represented by indefinite binary quadratic form Noam found that a (positive) odd prime $p$ is represented by the indefinite form $x^2 + 13 x y - 9 y^2$ if and only if $$ x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1 $$ has a root $\pmod p;$ indeed, in that case the polynomial factors into linear factors, distinct if $p > 5.$

Now, there is the possibility (I am not sure at all) of finding a polynomial $f(x)$ of degree 4 in the sense of Cox, page 180, Theorem 9.2: if an odd prime $p$ does not divide $205,$ then we have an integral expression $p = s^2 + 13 s t - 9 t^2$ if and only if $(205|p) = 1$ and $f(x) \equiv 0 \pmod p$ has a solution.

After looking again at Kronecker's result on page 88, $p = x^2 + 31 y^2$ if and only if $$ (x^3 - 10x)^2 + 31 (x^2 -1)^2 \equiv 0 \pmod p $$ has a root, compared with the cubic (from Hudson and Williams 1991) $(-31|p) = 1$ and $x^3 + x + 1$ factors completely, I looked and found this: if $u = x^4 + x^2 + 2$ and $v = x^2 + 3,$ we get $$ \color{magenta}{ u^2 + 13 u v - 9 v^2 \; \; = \; \; x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1}. $$ Despite understanding none of this, I think that easy identity means something.

So, there is the question, can we go from degree 8 with no congruence conditions, down to degree 4 with $(205|p) = 1?$

Extra: here is Franz's answer, all those years ago. Note the use of the word "compositum." People don't use a word like that unless they really mean it.

In his lovely answer at Positive primes represented by indefinite binary quadratic form Noam found that a (positive) odd prime $p$ is represented by the indefinite form $x^2 + 13 x y - 9 y^2$ if and only if $$ x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1 $$ has a root $\pmod p;$ indeed, in that case the polynomial factors into linear factors, distinct if $p > 5.$

Now, there is the possibility (I am not sure at all) of finding a polynomial $f(x)$ of degree 4 in the sense of Cox, page 180, Theorem 9.2: if an odd prime $p$ does not divide $205,$ then we have an integral expression $p = s^2 + 13 s t - 9 t^2$ if and only if $(205|p) = 1$ and $f(x) \equiv 0 \pmod p$ has a solution.

After looking again at Kronecker's result on page 88, $p = x^2 + 31 y^2$ if and only if $$ (x^3 - 10x)^2 + 31 (x^2 -1)^2 \equiv 0 \pmod p $$ has a root, compared with the cubic (from Hudson and Williams 1991) $(-31|p) = 1$ and $x^3 + x + 1$ factors completely, I looked and found this: if $u = x^4 + x^2 + 2$ and $v = x^2 + 3,$ we get $$ \color{magenta}{ u^2 + 13 u v - 9 v^2 \; \; = \; \; x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1}. $$ Despite understanding none of this, I think that easy identity means something.

So, there is the question, can we go from degree 8 with no congruence conditions, down to degree 4 with $(205|p) = 1?$

Extra: here is Franz's answer, all those years ago. Note the use of the word "compositum." People don't use a word like that unless they really mean it.

In his lovely answer at Positive primes represented by indefinite binary quadratic form Noam found that a (positive) odd prime $p$ is represented by the indefinite form $x^2 + 13 x y - 9 y^2$ if and only if $$ x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1 $$ has a root $\pmod p;$ indeed, in that case the polynomial factors into linear factors, distinct if $p > 5.$

Now, there is the possibility (I am not sure at all) of finding a polynomial $f(x)$ of degree 4 in the sense of Cox, page 180, Theorem 9.2: if an odd prime $p$ does not divide $205,$ then we have an integral expression $p = s^2 + 13 s t - 9 t^2$ if and only if $(-205|p) = 1$ and $f(x) \equiv 0 \pmod p$ has a solution.

After looking again at Kronecker's result on page 88, $p = x^2 + 31 y^2$ if and only if $$ (x^3 - 10x)^2 + 31 (x^2 -1)^2 \equiv 0 \pmod p $$ has a root, compared with the cubic (from Hudson and Williams 1991) $(-31|p) = 1$ and $x^3 + x + 1$ factors completely, I looked and found this: if $u = x^4 + x^2 + 2$ and $v = x^2 + 3,$ we get $$ u^2 + 13 u v - 9 v^2 \; \; = \; \; x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1. $$ Despite understanding none of this, I think that easy identity means something.

So, there is the question, can we go from degree 8 with no congruence conditions, down to degree 4 with $(-205|p) = 1?$

Extra: here is Franz's answer, all those years ago. Note the use of the word "compositum." People don't use a word like that unless they really mean it.

In his lovely answer at Positive primes represented by indefinite binary quadratic form Noam found that a (positive) odd prime $p$ is represented by the indefinite form $x^2 + 13 x y - 9 y^2$ if and only if $$ x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1 $$ has a root $\pmod p;$ indeed, in that case the polynomial factors into linear factors, distinct if $p > 5.$

Now, there is the possibility (I am not sure at all) of finding a polynomial $f(x)$ of degree 4 in the sense of Cox, page 180, Theorem 9.2: if an odd prime $p$ does not divide $205,$ then we have an integral expression $p = s^2 + 13 s t - 9 t^2$ if and only if $(205|p) = 1$ and $f(x) \equiv 0 \pmod p$ has a solution.

After looking again at Kronecker's result on page 88, $p = x^2 + 31 y^2$ if and only if $$ (x^3 - 10x)^2 + 31 (x^2 -1)^2 \equiv 0 \pmod p $$ has a root, compared with the cubic (from Hudson and Williams 1991) $(-31|p) = 1$ and $x^3 + x + 1$ factors completely, I looked and found this: if $u = x^4 + x^2 + 2$ and $v = x^2 + 3,$ we get $$ \color{magenta}{ u^2 + 13 u v - 9 v^2 \; \; = \; \; x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1}. $$ Despite understanding none of this, I think that easy identity means something.

So, there is the question, can we go from degree 8 with no congruence conditions, down to degree 4 with $(205|p) = 1?$

Extra: here is Franz's answer, all those years ago. Note the use of the word "compositum." People don't use a word like that unless they really mean it.

In his lovely answer at Positive primes represented by indefinite binary quadratic form Noam found that a (positive) odd prime $p$ is represented by the indefinite form $x^2 + 13 x y - 9 y^2$ if and only if $$ x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1 $$ has a root $\pmod p;$ indeed, in that case the polynomial factors into linear factors, distinct if $p > 5.$

Now, there is the possibility (I am not sure at all) of finding a polynomial $f(x)$ of degree 4 in the sense of Cox, page 180, Theorem 9.2: if an odd prime $p$ does not divide $205,$ then we have an integral expression $p = s^2 + 13 s t - 9 t^2$ if and only if $(-205|p) = 1$ and $f(x) \equiv 0 \pmod p$ has a solution.

After looking again at Kronecker's result on page 88, $p = x^2 + 31 y^2$ if and only if $$ (x^3 - 10x)^2 + 31 (x^2 -1)^2 \equiv 0 \pmod p $$ has a root, comapared with the cubic (from Hudson and Williams 1991) $(-31|p) = 1$ and $x^3 + x + 1$ factors completely, I looked and found this: if $u = x^4 + x^2 + 2$ and $v = x^2 + 3,$ we get $$ u^2 + 13 u v - 9 v^2 = x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1. $$ Despite understanding none of this, I think that easy identity means something.

So, there is the question, can we go from degree 8 with no congruence conditions, down to degree 4 with $(-205|p) = 1?$

In his lovely answer at Positive primes represented by indefinite binary quadratic form Noam found that a (positive) odd prime $p$ is represented by the indefinite form $x^2 + 13 x y - 9 y^2$ if and only if $$ x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1 $$ has a root $\pmod p;$ indeed, in that case the polynomial factors into linear factors, distinct if $p > 5.$

Now, there is the possibility (I am not sure at all) of finding a polynomial $f(x)$ of degree 4 in the sense of Cox, page 180, Theorem 9.2: if an odd prime $p$ does not divide $205,$ then we have an integral expression $p = s^2 + 13 s t - 9 t^2$ if and only if $(-205|p) = 1$ and $f(x) \equiv 0 \pmod p$ has a solution.

After looking again at Kronecker's result on page 88, $p = x^2 + 31 y^2$ if and only if $$ (x^3 - 10x)^2 + 31 (x^2 -1)^2 \equiv 0 \pmod p $$ has a root, compared with the cubic (from Hudson and Williams 1991) $(-31|p) = 1$ and $x^3 + x + 1$ factors completely, I looked and found this: if $u = x^4 + x^2 + 2$ and $v = x^2 + 3,$ we get $$ u^2 + 13 u v - 9 v^2 \; \; = \; \; x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1. $$ Despite understanding none of this, I think that easy identity means something.

So, there is the question, can we go from degree 8 with no congruence conditions, down to degree 4 with $(-205|p) = 1?$

Extra: here is Franz's answer, all those years ago. Note the use of the word "compositum." People don't use a word like that unless they really mean it.