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You want the idoneal numbers, http://oeis.org/A000926 and http://en.wikipedia.org/wiki/Idoneal_number

See also pages 81-82 in Duncan A. Buell, Binary Quadratic Forms

Depending what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.

EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).

Either way, there is a monic irreducible polynomial $f_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.

EDIT TOOO: What you want is Lemma 3.10 on page 333 of LIU_WILLIAMS Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994. I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking... Yes, this is correct. Theorem 4.1 on the same page, Table right below it. There is a bit of work showing that one root and $p \equiv 1 \pmod 8$ actually shows four linear factors.

You want the idoneal numbers, http://oeis.org/A000926 and http://en.wikipedia.org/wiki/Idoneal_number

See also pages 81-82 in Duncan A. Buell, Binary Quadratic Forms

Depending on what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.

EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).

Either way, there is a monic irreducible polynomial $f_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.

EDIT TOOO: What you want is Lemma 3.10 on page 333 of LIU_WILLIAMS Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994. I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking... Yes, this is correct. Theorem 4.1 on the same page, Table right below it. There is a bit of work showing that one root and $p \equiv 1 \pmod 8$ actually shows four linear factors.

You want the idoneal numbers, http://oeis.org/A000926 and http://en.wikipedia.org/wiki/Idoneal_number

See also pages 81-82 in Duncan A. Buell, Binary Quadratic Forms

Depending what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.

EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).

Either way, there is a monic irreducible polynomial $f_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.

EDIT TOOO: What you want is Lemma 3.10 on page 333 of LIU_WILLIAMS Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994. I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking... Yes, this is correct. Theorem 4.1 on the same page, Table right below it. There is a bit of work showing that one root and $p \equiv 1 \pmod 8$ actually shows four linear factors.

You want the idoneal numbers, http://oeis.org/A000926 and http://en.wikipedia.org/wiki/Idoneal_number

See also pages 81-82 in Duncan A. Buell, Binary Quadratic Forms

Depending what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.

EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).

Either way, there is a monic irreducible polynomial $f_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.

EDIT TOOO: What you want is Lemma 3.10 on page 333 of LIU_WILLIAMS Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994. I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking... Yes, this is correct. Theorem 4.1 on the same page, Table right below it. There is a bit of work showing that one root and $p \equiv 1 \pmod 8$ actually shows four linear factors.

You want the idoneal numbers, http://oeis.org/A000926 and http://en.wikipedia.org/wiki/Idoneal_number

See also pages 81-82 in Duncan A. Buell, Binary Quadratic Forms

Depending what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.

EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).

Either way, there is a monic irreducible polynomial $f_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.

EDIT TOOO: What you want is Lemma 3.10 on page 333 of LIU_WILLIAMS Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994. I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking...

You want the idoneal numbers, http://oeis.org/A000926 and http://en.wikipedia.org/wiki/Idoneal_number

See also pages 81-82 in Duncan A. Buell, Binary Quadratic Forms

Depending what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.

EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).

Either way, there is a monic irreducible polynomial $f_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.

EDIT TOOO: What you want is Lemma 3.10 on page 333 of LIU_WILLIAMS Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994. I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking... Yes, this is correct. Theorem 4.1 on the same page, Table right below it. There is a bit of work showing that one root and $p \equiv 1 \pmod 8$ actually shows four linear factors.