Fermat's 2 Square-Like Results from Minkowski Lattice Proofs 
Minkowski's Convex Body Theorem for lattices in the plane: Suppose $\mathfrak{L}$ is a lattice in $\mathbf{R}^2$ defined as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$, where $\vec{v_1}$ and $\vec{v_2}$ are linearly independent. Suppose $d$ is the area of a fundamental parallelogram of $\mathfrak{L}$. Then, if $\mathcal{S}$ is a convex and origin-symmetric region with $Area(\mathcal{S})>4d$, then $\mathcal{S}$ contains some point $q\neq 0$ such that $q\in\mathfrak{L}$.

There is a particularly nice proof of Fermat's two square theorem - an odd prime $p$ is expressible as $x^2+y^2$ for $x,y\in\mathbf{Z}$ $\iff$ $p\equiv 1\pmod{4}$ - using this fact. Here is my proof below:

Choose $a\in\mathbf{Z}$ such that $p|a^2+1$. Now, let $\mathfrak{L}$ be a lattice in $\mathbf{R}^2$ defined as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$ where $\vec{v_1}=(a,1)$ and $\vec{v_2}=(p,0)$. It is easy to see that a fundamental parallelogram $\mathfrak{F}$ of $\mathfrak{L}$ has area $p$. Now, suppose $(x,y)\in\mathfrak{L}$. Then we have $x=ma+np, y=m$, so \begin{align*}x^2+y^2 &= (ma+np)^2 + m^2 \\ &= m^2a^2+2manp+n^2p^2+m^2\\ &= m^2(a^2+1) + 2manp + n^2p^2 \\ &\equiv m^2(a^2+1)\pmod{p} \\ &\equiv 0\pmod{p} \end{align*} due to our choice of $a$. Now, let $\mathfrak{C}$ be an $O$-symmetric circle with radius $\sqrt{2p}$, so $\mathfrak{C}=\{(x,y)\in\mathbf{R}^2 : x^2+y^2<2p\}$ Then $$Area(\mathfrak{C})=2p\pi > 4p = 4\cdot Area(\mathfrak{F}).$$ Thus, by Minkowski, $\exists$ a lattice point $(j,k)\in\mathfrak{C}\setminus (0,0)$. By the above work, $p|j^2+k^2$, and by definition of $\mathfrak{C}$, $j^2+k^2<2p$. Moreover, $j^2+k^2$ is positive, so $0<j^2+k^2<2p$, but in order for $j^2+k^2$ to be divisible by $p$, we must have $p=j^2+k^2$, so we are done.

Are there any other results that can be derived in a similar manner? Specifically, what primes can be expressed as $x^2+2y^2$? $x^2+3y^2$? Any ideas are greatly appreciated.

Unfortunately I'm answering my own question, but I was able to prove that if $p\equiv 1\pmod{8}$ or $p\equiv 3\pmod{8}$, then $p$ is expressible as $x^2+2y^2$ for $x,y\in\mathbf{Z}$, and if $p\equiv 1\pmod{3}$, then $p$ is expressible as $x^2+3y^2$ for $x,y\in\mathbf{Z}$. For posterity, here it is.
Lemma: $p\equiv 1\pmod{8}$ or $p\equiv 3\pmod{8}\iff p|a^2+2$ for some $a\in\mathbf{Z}$.
Proof: The latter statement is equivalent to $\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=1$. It is not hard to see that this holds if and only if $p$ is of the shape $8k+1$ or $8k+3$ due to CRT. $\Box$
By our lemma, we can choose $a\in\mathbf{Z}$ such that $p|a^2+2$. Define a lattice $\mathfrak{L}\in\mathbf{R}^2$ as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$ where $\vec{v_1}=(a,1)$ and $\vec{v_2}=(p,0)$. The fundamental parallelogram $\mathfrak{F}$ of $\mathfrak{L}$ has area $p$. Let $(x,y)\in\mathfrak{L}$. Then $x=ma+np$ and $y=m$, so \begin{align*}x^2+2y^2 &= (ma+np)^2+2m^2 \\ &= m^2a^2+2manp+n^2p^2+2m^2 \\ &= m^2(a^2+2)+2manp+n^2p^2 \\ &\equiv m^2(a^2+2)\pmod{p}\\ &\equiv 0\pmod{p}\end{align*} due to our choice of $a$. Now, let $\mathfrak{E}$ be a convex, origin symmetric ellipse with semi-major axis length $\sqrt{2p}$ and semi-minor axis length $\sqrt{p}$, so $\mathfrak{E}=\{(x,y)\in\mathbf{R}^2:x^2+2y^2<2p\}$. Then, we have $$Area(\mathfrak{E})=p\pi\sqrt{2}>4p=4\cdot Area(\mathfrak{F}).$$ Thus, by Minkowski there exists a lattice point $(j,k)\in\mathfrak{E}\setminus(0,0).$ By the above work, $p|j^2+2k^2$ and by definition of $\mathfrak{E}$, $j^2+2k^2<2p$. $j^2+2k^2$ is obviously positive, so $0<j^2+2k^2<2p$, so in order for $p|j^2+2k^2$ we must have $p=j^2+2k^2$, which completes the proof.
UPDATE: I have included my proof of the proposition for primes expressible as $x^2+3y^2$ for $x,y\in\mathbf{Z}$.
Lemma: $p\equiv 1\pmod{3} \iff p|a^2+3$ for some $a\in\mathbf{Z}$.
Proof: The latter statement is equivalent to $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=1$. It is well known that this occurs if and only if $p\equiv 1\pmod{6}$. The proof is a basic application of the Quadratic Reciprocity law and the Chinese Remainder Theorem. Since all primes congruent to $1$ modulo $6$ are exactly those congruent to $1$ modulo $3$, we are done. $\Box$
By our lemma, we can choose $a\in\mathbf{Z}$ such that $p|a^2+3$. Define a lattice $\mathfrak{L}\in\mathbf{R}^2$ as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$ where $\vec{v_1}=(a,1)$ and $\vec{v_2}=(p,0)$. The fundamental parallelogram $\mathfrak{F}$ of $\mathfrak{L}$ has area $p$. Let $(x,y)\in\mathfrak{L}$. Then $x=ma+np$ and $y=m$, so \begin{align*}x^2+3y^2 &= (ma+np)^2+3m^2 \\ &= m^2a^2+2manp+n^2p^2+3m^2 \\ &= m^2(a^2+3)+2manp+n^2p^2 \\ &\equiv m^2(a^2+3)\pmod{p}\\ &\equiv 0\pmod{p}\end{align*} due to our choice of $a$. Now, let $\mathfrak{E}$ be a convex, origin symmetric ellipse with semi-major axis length $\sqrt{3p}$ and semi-minor axis length $\sqrt{p}$, so $\mathfrak{E}=\{(x,y)\in\mathbf{R}^2:x^2+3y^2<3p\}$. Then, we have $$Area(\mathfrak{E})=p\pi\sqrt{3}>4p=4\cdot Area(\mathfrak{F}).$$ Thus, by Minkowski's Convex Body Theorem, there exists a lattice point $(j,k)\in\mathfrak{E}\setminus(0,0).$ By the above work, $p|j^2+3k^2$ and by definition of $\mathfrak{E}$, $j^2+3k^2<3p$. $j^2+3k^2$ is obviously positive, so $0<j^2+3k^2<3p$, so in order for $p|j^2+3k^2$ we must have $p=j^2+3k^2$ or $2p=j^2+3k^2$. In order to show that the latter possibility is impossible, consider the equation modulo $3$. We get, since $p\equiv 1\pmod{3}$, $j^2+3k^2\equiv j^2\equiv 2p\equiv 2\pmod{3}$, but this is impossible since the only quadratic residues modulo $3$ are $0$ and $1$. Hence proved. $\Box$
