Consider the quadratic form $x^2 + 2y^2$ over the ring $\mathbb{Z}[i]$. It is irreducible, so I wanted to know which primes could be represented by this quadratic form.

$$ \mathfrak{p} = x^2 + 2y^2 $$

There's two slightly different questions here. One about the integer $\mathfrak{p} \in \mathbb{Z}[i]$ and one about the ideals $(\mathfrak{p}) \subseteq \mathbb{Z}[i]$. This is still a PID so I can write just one element.

This hopefully in analogy to Fermat's theorem on primes as the sum of two squares, where it is solved by a congruence condition. The analogous question for $x^2 + y^2 = (x+iy)(x-iy)$ is degenerate.

  • $\begingroup$ Do you want primes which do not split over $\mathbb{Z}[i]$, that is, the rational primes $p \equiv 3 \pmod{4}$? $\endgroup$ – Stanley Yao Xiao Jul 6 '18 at 15:27
  • $\begingroup$ @StanleyYaoXiao $\mathfrak{p}=1+2i$ is prime, and now I can attempt to solve $\mathfrak{p} = a^2 + b^2 $ with $a,b \in \mathbb{Z}[i]$. $\endgroup$ – john mangual Jul 6 '18 at 15:38
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    $\begingroup$ The relevant number field ${\bf Q}(i,\sqrt{-2})$ is isomorphic with the 8th cyclotomic field, which has trivial class group <lmfdb.org/NumberField/>. So there should be a necessary and sufficient congruence condition here too, maybe as simple as $\mathfrak p$ having norm $1 \bmod 8$. $\endgroup$ – Noam D. Elkies Jul 6 '18 at 23:57

Here is a partial answer. Hope someone out there would fill in the remaining details. In summary, using the OP's notations (in particular $x,y\in {\mathbb Z}[i]$), the case for $p=2$ is not possible, i.e. $\pm (1+i)$ is not of the form $x^2+2y^2$. If $p\equiv 1~({\rm mod~}4)$, then necessarily $p:=\frak{ p}\overline {\frak {p}}$ must satisfy $$p\equiv 1~({\rm mod~}8).$$ For $p\equiv 3~({\rm mod~}4),$ it is always possible to write $p=x^2+2y^2$.

For example, in the situation $p\equiv 1~({\rm mod~}8)$, the first few cases of $\frak{p}$ are as follows (one takes complex conjugate to get the representation for $\overline{\frak p}$): $$p=17=\frak{p}\overline{\frak{p}}~{\rm with~}{\frak{p}}=1+4i=1^2+2(1+i)^2$$ $$p=41=\frak{p}\overline{\frak{p}}~{\rm with~}{\frak{p}}=5+4i=(2+i)^2+2\cdot 1^2$$ $$p=73=\frak{p}\overline{\frak{p}}~{\rm with~}{\frak{p}}=-3+8i=(1+2i)^2+2(1+i)^2$$ $$p=89=\frak{p}\overline{\frak{p}}~{\rm with~}{\frak{p}}=5+8i=(3+2i)^2+2(1-i)^2$$ $$p=97=\frak{p}\overline{\frak{p}}~{\rm with~}{\frak{p}}=9+4i=3^2+2(1+i)^2.$$ The reasoning (incomplete when $p\equiv 1~({\rm mod~}8)$) is as follows: Consider first the case when $p\equiv 1~({\rm mod~}4)$. Denoting $x=a+bi,y=c+di,a,b,c,d\in {\mathbb Z}$ with $p={\frak{p}}\overline{\frak{p}}=(x^2+2y^2)(\overline{x}^2+2\overline{y}^2).$ It follows that $$p=(a^2+b^2)^2+4(c^2+d^2)^2+4[(ac+bd)^2-(bc-ad)^2],$$ from which by some parity checking, it is easy to see that necessarily $p\equiv 1~({\rm mod~}8).$

Now it remains to check the cases $p=2$ and $p\equiv 3~({\rm mod~}4).$ For $p=2$, one needs to have $$1\pm i=x+2y^2=(a+bi)^2+2(c+di)^2=[a^2-b^2+2(c^2-d^2)]+2i[ab+2cd],$$ which is absurd.

For $p\equiv 3~({\rm mod~}4)$, $p$ is prime, so we need to solve the condition $$p=[a^2-b^2+2(c^2-d^2)]+2i[ab+2cd],$$ hence $$p=a^2-b^2+2(c^2-d^2)~{\rm and~}ab=-2cd.$$ Two subcases arise.

Subcase 1: $p\equiv 3~({\rm mod~}8)$. Here one just solves from the above condition for $p=a^2+2c^2$ (letting $b=d=0$), which is solvable, since $$p=u^2+2v^2\Leftrightarrow p\equiv 1,3~({\rm mod~}8).$$ (See D. Cox's famous book "Primes of the forms $x^2+ny^2$.")

Subcase 2: $p\equiv -1~({\rm mod}~8)$. This time, use the relation $p=a^2-2d^2$ (letting $b=c=0$). This is possible by the following known result: $$p\equiv \pm 1~({\rm mod~}8)\Leftrightarrow p=u^2-2v^2$$ (See F. Lemmermeyer's notes here. Link to be added.) Now it boils down to solving the remaining details:

Problem. Show that when $p\equiv 1~({\rm mod~}8),$ the equation $${\frak{p}}=x^2+2y^2$$ can be solved, where $p=\frak{p}\overline{\frak{p}}.$ This may require the comments from Noam D. Elkies.


Just elaborating on the very nice comments and answer by Cherng-tiao Perng. As Noam Elkies said, $$ \mathbb{Q}[i,\sqrt{-2}]=\mathbb{Q}[\exp(2\pi i/8)]. $$ As is well known, the Galois group of the $8$-th cyclotomic field over $\mathbb{Q}$ is isomorphic to $(\mathbb{Z}/8\mathbb{Z})^{*}=\{1,3,5,7\}$. Let $\sigma_j$ be the element in $Gal(\mathbb{Q}[\exp(2\pi i/8)]/\mathbb{Q})$ that sends $\exp(2\pi i/8)$ to $\exp(2\pi ij/8)$, $j\in \{1,3,5,7\}$.

As is well known, $\sigma_5$ fixes $i$, sends $\sqrt{2}$ to $-\sqrt{2}$, $\sqrt{-2}$ to $-\sqrt{-2}$. $\sigma_3$ fixes $\sqrt{-2}$, sends $i$ to $-i$, $\sqrt{2}$ to $-\sqrt{2}$. $\sigma_7$ fixes $\sqrt{2}$, sends $i$ to $-i$, $\sqrt{-2}$ to $-\sqrt{-2}$.

Let $x=a+ib$, $y=c-id$. If $\mathfrak{p}\in\mathbb{Z}[i]$ may be written as $$ \mathfrak{p}=x^2+2y^2=(x+y\sqrt{-2})(x-y\sqrt{-2})=(a+ib+\sqrt{-2}c+\sqrt{2}d)(a+ib-\sqrt{-2}c-\sqrt{2}d), $$ then $$\mathfrak{p}=(a+ib+\sqrt{-2}c+\sqrt{2}d)\sigma_5((a+ib+\sqrt{-2}c+\sqrt{2}d)).$$

Since the primes $\mathfrak{p}$ in $\mathbb{Z}[i]$ are rational primes $p=3,7\bmod 8$ or primes with norms that are rational primes $p=1,5\bmod 8$, we consider these cases.

Suppose $p=1,5\bmod 8$ so that $p=\mathfrak{p}\mathfrak{\bar{p}}$. Since $\mathfrak{\bar{p}}=\sigma_3(\mathfrak{p})$, then if $$ \mathfrak{p}=(a+ib+\sqrt{-2}c+\sqrt{2}d)\sigma_5((a+ib+\sqrt{-2}c+\sqrt{2}d)), $$ then $$ \mathfrak{\bar{p}}=\sigma_3((a+ib+\sqrt{-2}c+\sqrt{2}d))\sigma_7((a+ib+\sqrt{-2}c+\sqrt{2}d)) $$ and $$ p=\prod_{j\in\{1,3,5,7\}}\sigma_j((a+ib+\sqrt{-2}c+\sqrt{2}d)). $$ As is well known, for odd primes $p$, this only happens if $p=1\bmod 8$.

As $\{a_0+a_1i+a_2\sqrt{2}+a_3\sqrt{-2}| a_0,a_1,a_2,a_3\in\mathbb{Z}\}$ is a subring of $\mathbb{Z}[\exp(2\pi i/8)]$, i.e., $$ a_0+a_1i+a_2\sqrt{2}+a_3\sqrt{-2}=a_0+a_1i+(a_2+a_3)\exp(2\pi i/8)+(a_3-a_2)\exp(2\pi i3/8), $$ as is pointed out in the other answer, we have to also check, when $p=1\bmod 8$ with $p=\mathfrak{p}\mathfrak{\bar{p}}$, where $\mathfrak{p}$ is an element and not an ideal, which of $\mathfrak{p}^{\prime}\in\mathfrak{p}\{1,-1,i,-i\}$ may be written as $$ \mathfrak{p}^{\prime}=\prod_{j\in\{1,5\}}\sigma_j(a_0+a_1i+a_2\sqrt{2}+a_3\sqrt{-2}), $$ and as the other answer indicates, and we may verify, for example, using PARI/GP


[Mod(Mod(-1, y^2 + 1)*x + Mod(y + 2, y^2 + 1), x^2 - 2), 1]


[Mod(Mod(11/2*y - 3/2, y^2 + 1)*x + Mod(8*y - 2, y^2 + 1), x^2 - 2), 1]

and also prove using congruences, that for such situations it only happens when $\mathfrak{p}^{\prime}$ is of the form $a+ib$, $a,b\in\mathbb{Z}$, $a$ odd, $b=0\bmod 4$.

Suppose that $p=3\bmod 8$. Then there are $x,y\in\mathbb{Z}$ so that $$ p=x^2+2y^2. $$

Suppose that $p=7\bmod 8$. Then there is $x\in\mathbb{Z}$, $y\in i\mathbb{Z}$ so that $$ p=x^2+2y^2. $$

  • $\begingroup$ Should have adjoined $\sqrt{-2}$ instead of $\sqrt{2}$ in the PARI/GP code. But anyway, same effect. $\endgroup$ – abc Jul 18 '18 at 4:45

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