A few days ago Serre told me about some modest improvements to the proof, based on Weil's book Number theory: an approach through history from Hammurapi to Legendre and on a 1998 letter from Deligne to Serre; I will paraphrase these below.

According to Weil (p. 292), the ``magical'' argument is due to an amateur mathematican: L. Aubry, Sphinxe-Oedipe 7 (1912), 81--84. Here is a generalization that allows for a clearer proof.

Lemma: Let $f = f_2+f_1+f_0 \in \mathbf{Z}[x_1,\ldots,x_n]$, where $f_i$ is homogeneous of degree $i$. Suppose that for every $x \in \mathbf{Q}^n-\mathbf{Z}^n$, there exists $y \in \mathbf{Z}^n$ such that $0<|f_2(x-y)|<1$. If $f$ has a zero in $\mathbf{Q}^n$, then it has a zero in $\mathbf{Z}^n$.

Proof: If $x=(x_1,\ldots,x_n) \in \mathbf{Q}^n$, let $\operatorname{den}(x)$ denote the lcm of the denominators of the $x_i$. By iteration, the following claim suffices: If $x \in \mathbf{Q}^n - \mathbf{Z}^n$ and $y \in \mathbf{Z}^n$ satisfy $0<|f_2(x-y)|<1$, and the line $L$ through $x$ and $y$ intersects $f=0$ in $x,x'$, then $\operatorname{den}(x')<\operatorname{den}(x)$. By restricting to $L$ and choosing a coordinate $t$ on it taking the value $0$ at $y$ and integer values exactly on $L \cap \mathbf{Z}^n$, we reduce to proving the following: given $f(t)=At^2+Bt+C \in \mathbf{Z}[t]$ with zeros $x,x' \in \mathbf{Q}$ such that $0<|Ax^2|<1$, we have $\operatorname{den}(x')<\operatorname{den}(x)$. Proof: $0<|Ax^2|<1$ implies $0<|A|<\operatorname{den}(x)^2$, and we have $xx'=C/A$, so $\operatorname{den}(x) \operatorname{den}(x') \le |A| < \operatorname{den}(x)^2$, so $\operatorname{den}(x')<\operatorname{den}(x)$.

A few days ago Serre told me about some modest improvements to the proof, based on Weil's book Number theory: an approach through history from Hammurapi to Legendre and on a 1998 letter from Deligne to Serre; I will paraphrase these below.

According to Weil (p. 292), the ``magical'' argument is due to an amateur mathematican: L. Aubry, Sphinxe-Oedipe 7 (1912), 81--84. Here is a generalization that allows for a clearer proof.

Lemma: Let $f = f_2+f_1+f_0 \in \mathbf{Z}[x_1,\ldots,x_n]$, where $f_i$ is homogeneous of degree $i$. Suppose that for every $x \in \mathbf{Q}^n-\mathbf{Z}^n$, there exists $y \in \mathbf{Z}^n$ such that $0<|f_2(x-y)|<1$. If $f$ has a zero in $\mathbf{Q}^n$, then it has a zero in $\mathbf{Z}^n$.

Proof: If $x=(x_1,\ldots,x_n) \in \mathbf{Q}^n$, let $\operatorname{den}(x)$ denote the lcm of the denominators of the $x_i$. By iteration, the following claim suffices: If $x \in \mathbf{Q}^n - \mathbf{Z}^n$ and $y \in \mathbf{Z}^n$ satisfy $0<|f_2(x-y)|<1$, and the line $L$ through $x$ and $y$ intersects $f=0$ in $x,x'$, then $\operatorname{den}(x')<\operatorname{den}(x)$. By an affine change of variable over $\mathbf{Z}$, we may assume that $y$ is $0$ and that $L$ is the $x_1$-axis. By restricting to $L$, we reduce to proving the following: given $f(t)=At^2+Bt+C \in \mathbf{Z}[t]$ with zeros $x,x' \in \mathbf{Q}$ such that $0<|Ax^2|<1$, we have $\operatorname{den}(x')<\operatorname{den}(x)$. Proof: Factor $f$ over $\mathbf{Z}$ as $E(Dt-N)(D't-N')$ with $x=N/D$ and $x'=N'/D'$ in lowest terms. Then $0<|Ax^2|<1$ implies $0<|A|<D^2$. On the other hand, $DD'$ divides $EDD'=A$, so $DD' \le |A| < D^2$. Hence $D'<D$.