In his lovely answer at [Positive primes represented by indefinite binary quadratic form](https://mathoverflow.net/questions/171807/positive-primes-represented-by-indefinite-binary-quadratic-form/171810#171810) 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](https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind9709&L=NMBRTHRY&P=R352&1=NMBRTHRY&9=A&J=on&d=No%20Match;Match;Matches&z=4), all those years ago. Note the use of the word "compositum." People don't use a word like that unless they really mean it. 



Positive Primes represented by an indefinite binary form, reducing poly degree from 8 to 4
I think the answer to the question as asked is no, but with $(205|p)=1$ it is yes, for then taking subfields of the Elkies field, look at $x^4 +x^3 + 3x^2 + 2x + 4$.


