Exercise 3.5.10 (Prime numbers represented by $x^2 + 5y^2$)

Proof.

(a)

Suppose that the prime $p$ is represented by the form $x^2+5y^2$. By Problem 9, $4p$ is a square modulo $|d|=20$, so $4p\equiv a^2(\mathit{mod}20)$ for some integer $a$ (this is obvious here: if $p=x_0^2+5y_0^2$ for some integers $x_0,y_0$, then $4p=4x_0^2+20y_0^2\equiv {(2x_0)}^2=a^2(\mathit{mod}20)$, where $a=2x_0$).

Then $4\mid a^2$, thus $2\mid a$, so $a=2b$ for some integer $b$, and $4p\equiv 4b^2(\mathit{mod}20)$, therefore

$$p\equiv b^2(\mathit{mod}5),$$

where $5$ is prime. This implies that

$$\left(\frac{p}{5}\right)=0\text{ or }1.$$

Suppose now that $p$ is represented by $2x^2+2\mathit{xy}+3y^2$. By the same Problem 9, $8p$ is a square modulo $20$, so $8p\equiv c^2(\mathit{mod}20)$. Therefore $c$ is even, so $c=2d$ for some integer $d$, hence

$$2p\equiv d^2(\mathit{mod}5).$$

Note that $5$ is not represented by the form $2x^2+2\mathit{xy}+3y^2$, otherwise $5=2x_0^2+2x_0{y}_0+3y_0^2$ for some integers $x_0,y_0$. Then $10={(2x_0+y_0)}^2+5y_0^2\ge 5y_0^2$, thus $y_0^2\le 2$, so $y_0\in \{0,1,-1\}$. If $y_0=\pm 1$, then ${(2x_0+y_0)}^2=5$, but $5$ is not a perfect square, and if $y=0$, then $10={(2x_0)}^2$, and this is also impossible. This shows that $p\ne 5$, so $\left(\frac{2p}{5}\right)\ne 0$.

Moreover $2p\equiv d^2(\mathit{mod}5)$, thus

$$\left(\frac{2p}{5}\right)=1.$$

Then $1=\left(\frac{2}{5}\right)\left(\frac{p}{5}\right)=-\left(\frac{p}{5}\right)$, thus

$$\left(\frac{p}{5}\right)=-1.$$

To conclude,

$$\begin{array}{cc}\exists <!--\; FUNCTION\; APPLICATION\; -->(x_0,y_0)\in {\mathbb{Z}}^2,p=x_0^2+5y_0^2\hfill & \Rightarrow \left(\frac{p}{5}\right)=0\text{ or }\left(\frac{p}{5}\right)=1,\hfill \\ \exists <!--\; FUNCTION\; APPLICATION\; -->(x_0,y_0)\in {\mathbb{Z}}^2,p=2x_0^2+2y_0^2+3y_0^2\hfill & \Rightarrow \left(\frac{p}{5}\right)=-1.\hfill \end{array}$$

(b)

By Problem 5, a prime $p$ is represented by $f_1(x,y)=x^2+5y^2$ or by $f_2(x,y)=2x^2+2y^2+3y^2$ if and only if $p=2$, $p=5$ or $p\equiv 1,3,7,9(\mathit{mod}20)$. By part (a), $p$ cannot be represented simultaneously by the two forms $f_1,f_2$.

• If $p=2$, then $p$ is represented by $f_2$, but not by $f_1$ by Problem 5.
• If $p=5$, then $p=5=f_1(0,1)$ is represented by $f_1$, but not by $f_2$ (see above).
• If $p\equiv 1,9(\mathit{mod}20)$, then $p\equiv 1,4(\mathit{mod}5)$ thus $\left(\frac{p}{5}\right)=1$. This shows that $p$ is not represented by $f_2$, hence by Problem 5, $p$ is represented by $f_1$.
• If $p\equiv 3,7(\mathit{mod}20)$, then $p\equiv 3,2(\mathit{mod}5)$, thus $\left(\frac{p}{5}\right)=-1$. This shows that $p$ is not represented by $f_1$, hence $p$ is represented by $f_2$.

In conclusion, a prime $p$ is represented by the form $x^2+5y^2$ if and only if $p=5$ or $p\equiv 1,9(\mathit{mod}20)$, and $p$ is represented by the form $2x^2+2\mathit{xy}+3y^2$ if and only if $p=2$ or $p\equiv 3,7(\mathit{mod}20)$.