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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.

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Wow. Costa Rica 2, Uruguay 1. Minute 58 –  Will Jagy Jun 14 at 20:17

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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 $x^4 +x^3 + 3x^2 + 2x + 4$.

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Ah! absolutely right, I put -205 out of habit, I mostly do positive binaries... –  Will Jagy Jun 14 at 20:39
    
Works perfectly. It is also $(x+1)^2 (x+2)^2 \pmod 5,$ then $(x+24)^2 (x^2 + 35 x + 2) \pmod {41}$ and 4 roots thereafter. So, to include 5, we can say $(205|p) \neq -1$ and all linear factors $\pmod p.$ –  Will Jagy Jun 14 at 21:01

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