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\hypersetup{pdfauthor={Roelof van Dijk},pdftitle={Gauss's class-number-one conjecture in 1808}}
\begin{document}
\begin{center}
{\Large\bfseries A proof Gauss might have found}\\[9pt]
Roelof van Dijk
\end{center}

\begin{abstract}
Ben Green's historical variant of Tao's recent tenth discriminant challenge~\cite{Tao} asks how early
Gauss's class-number-one conjecture could have been proved using mathematics then in print.
Through discussions with GPT-6 Astra and Opus 5.5, we construct a proof of Gauss's class-number-one conjecture~\cite{Gauss}
from ingredients published by 1808, specializing the Frobenius determinant argument~\cite{vanDijk}.
Counting values of the principal form and summing factorial valuations reduce the proof
to a residue sieve below~$17000$.
\end{abstract}

\begin{theorem}
The negative fundamental discriminants of class number one are
\[
 -3,\ -4,\ -7,\ -8,\ -11,\ -19,\ -43,\ -67,\ -163.
\]
\end{theorem}

\begin{proof}
Write the discriminant as $-q$ and suppose $h(-q)=1$.

The principal form~\cite{Gauss} is $f=(x+y/2)^2+qy^2/4$ if $q\equiv3\pmod4$ and
$f=x^2+qy^2/4$ if $4\mid q$.
Its values are squares or at least $q/4$. Call a prime $\ell$ inert if
$b^2\equiv-q\pmod{4\ell}$ has no solution. Otherwise $\left(\ell,b,\frac{b^2+q}{4\ell}\right)$ is primitive,
since $-q$ is fundamental, and is equivalent to $f$. Thus every non-inert prime is a value
$f(u,v)$ with $v\ne0$, and is at least $q/4$. In particular,
\begin{equation}\label{inert}
 \text{every prime }\ell<q/4\text{ is inert}.
\end{equation}
Assume $q>8$. Inertness of $2$ gives $q\equiv3\pmod8$.
Then $q$ is odd and squarefree, and must be prime: a composite odd squarefree integer has
a prime factor below $q/4$, whereas every prime factor of $q$ is non-inert.

Put $\omega=(1+\sqrt{-q})/2$ and $\mathcal O=\Z[\omega]$, with
$\omega^2=\omega-(q+1)/4$. Write $\bar z$ for the conjugate of $z$.
For every inert prime $\ell$ and $z\in\mathcal O$,
\begin{equation}\label{euler}
 z^\ell\equiv\bar z\pmod{\ell\mathcal O}.
\end{equation}
Indeed $(a+b\omega)^\ell\equiv a+b\omega^\ell$ by the binomial theorem and Fermat.
For odd $\ell$, Euler's criterion gives $(2\omega)^\ell\equiv1-\sqrt{-q}=2\bar\omega$.
Since $2^\ell\equiv2$ and $\ell$ is odd, $\omega^\ell\equiv\bar\omega$.
At $\ell=2$, the integer $(q+1)/4$ is odd, so $\omega^2\equiv\omega+1\equiv\bar\omega$.
For an odd inert $p$, the identity
$4(a^2+ab+\tfrac{q+1}4b^2)=(2a+b)^2+qb^2$ shows that the norm of $a+b\omega$ is prime to $p$ unless
$p\mid a,b$. Thus $\mathcal O/p\mathcal O$ is a field, with inverses $\bar z/(z\bar z)$, in which
$z^p=\bar z$.

We use~\eqref{euler} twice: first to prove a determinant nonzero modulo a large inert prime,
then to give it too many prime factors for its size.
For an inert prime $p$, the determinant will have order $pS$, where $S=\lfloor p/40\rfloor$.
After division by $Sp^2$, the logarithm of its norm will be at least $\log p$ minus an
absolute constant, and at most $\tfrac{21}{40}\log p+\tfrac14\log q$.
Choosing $q^3/(5\log q)<p\le q^3$ makes the difference grow like $\tfrac{47}{40}\log q$.
The principal-form count supplies $p$ and bounds the contribution of the non-inert primes.

Suppose $q\ge17000$, and let $N(x)$ count the non-inert primes at most $x$.
Choose their representations with $v>0$ and $u+v/2\ge0$. For $x>0$,
\[
 N(x)\le\sum_{1\le v\le2\sqrt{x/q}}
       \left(\sqrt{x-qv^2/4}+1\right)
 \le\frac{5x}{3\sqrt q}+2\sqrt{\frac xq}.
\]
The last inequality follows from $\sqrt{1-t^2}\le1-t^2/2$ and comparison with an integral.
For $\ell\le Y=q^3$,
\[
 \frac{\log\ell}{\ell}=\frac{\log Y}{Y}
       +\int_\ell^Y\frac{\log t-1}{t^2}\,dt.
\]
Adding over the non-inert primes and using $N(t)=0$ for $t\le q/4$ gives
\begin{equation}\label{scarce}
 \sum_{\substack{\ell\le q^3\\\ell\ \text{not inert}}}\frac{\log\ell}\ell
 \le\frac5{6\sqrt q}\bigl(8\log^2q+5\log q\bigr)+\frac{8\log q+8}{q}<5.2.
\end{equation}
The bound decreases for $q\ge17000$. At $17000$, use $\log q<9.75$ and $\sqrt q>130$.

Let $n=\lfloor q^3/2\rfloor$. Legendre's formula~\cite[Introduction, art.~XVI, pp.~8--9]{Legendre} gives
$v_\ell\binom{2n}n\le\log(2n)/\log\ell$: each summand
$\lfloor2n/\ell^j\rfloor-2\lfloor n/\ell^j\rfloor$ is $0$ or $1$, and vanishes for $\ell^j>2n$.
Thus every prime power exactly dividing $\binom{2n}n$ is at most $2n$.
There are at most $X/2+1$ primes up to $X=q^3/(5\log q)$.
If all the inert prime divisors of $\binom{2n}n$ were at most $X$, then
\[
 4^n/(2n+1)\le\tbinom{2n}n\le(2n)^{X/2+1+N(2n)}.
\]
The logarithms of the two bounds are greater than $0.69q^3$ and less than $0.68q^3$, respectively.
Hence there is an inert prime $p$ with
\[
 q^3/(5\log q)<p\le q^3,\qquad p>10^{11}.
\]
Put
\[
 m=\left\lceil\sqrt{\frac{p}{2\sqrt q}}\right\rceil,\qquad B=m\sqrt{q/2}.
\]
The elements $a+b\omega$ with $|b|\le m$ and $|2a+b|\le m\sqrt q$ number at least
$(2m+1)\lfloor m\sqrt q\rfloor\ge2\sqrt q\,m^2\ge p$, and have absolute value at most $B$.
The difference of two of them has norm at most $4B^2<p^2$, so they are distinct modulo $p$.
Choose $p$ of them, $z_1,\dots,z_p$. Then
\begin{equation}\label{height}
 \log B\le\tfrac12\log p+\tfrac14\log q-\log2+0.001.
\end{equation}

Match the $S$ successive $y$-degrees with the $S$ derivative orders in $x$.
The degrees provide room for factors $x^\ell-y$, while the derivatives cancel the added powers of $B$.
Form the square matrix $A$ with rows $(r,s)$, $0\le r<p$, $0\le s<S$, columns $(i,j)$,
$1\le i\le p$, $0\le j<S$, and entries
\[
 A_{(r,s),(i,j)}=\partial_x^j(x^ry^s)\big|_{(x,y)=(z_i,\bar z_i)}\in\mathcal O.
\]
These are ordinary derivatives. Dividing the $\ell$th derivative of $x^\ell$ by $\ell!$
would lose the factor $\ell$ needed below.
A relation between the rows modulo $p$, with coefficients $c_{rs}$, gives the polynomial
$G(x)=\sum c_{rs}x^{r+ps}$. Since $\bar z_i=z_i^p$ and $(x^{ps})'=0$ in characteristic $p$,
we have $G^{(j)}(z_i)=0$ for $j<S$.
As $S<p$, the $p$ distinct roots $z_i$ each have multiplicity at least $S$.
But $\deg G<pS$, so $G=0$. The exponents $r+ps$ are distinct, hence every $c_{rs}$ is zero.
Thus $\Delta=\det A\ne0$ modulo $p$, and $D=\Delta\bar\Delta$ is a positive integer.

Each entry has absolute value at most $p^jB^{r+s-j}$.
The derivative orders cancel the $s$-degrees:
\[
 \sum_{i,j}j=\sum_{r,s}s=\frac{pS(S-1)}2.
\]
The Leibniz expansion therefore gives
\[
 \begin{aligned}
 \log D&\le2\log((pS)!)+pS(p-1)\log B+pS(S-1)\log p\\
       &\le Sp^2\bigl(\log B+\tfrac1{40}\log p+10^{-6}\bigr),
 \end{aligned}
\]
since $S\le p/40$ and $4\log p/p<10^{-6}$.

Fix an inert prime $\ell<p$.
Replace row $(r,s)$ by the row of $x^{r-\ell h}(x^\ell-y)^hy^s$, where
$h=\min(\lfloor r/\ell\rfloor,S-1-s)$. The added monomials have smaller $x$-degree and $y$-degree
below $S$. Ordered by $x$-degree, the change of basis is triangular with diagonal entries one,
so $\Delta$ is unchanged. By~\eqref{euler} and Leibniz's rule each new entry lies in
$\ell^h\mathcal O$, because $z_i^\ell-\bar z_i\in\ell\mathcal O$ and every positive derivative of $x^\ell$
has coefficient divisible by $\ell$. Factoring $\ell^h$ from each row gives
$\Delta\in\ell^{\sum_{r,s}h}\mathcal O$, hence $\ell^{2\sum_{r,s}h}\mid D$,
and these integer divisibilities multiply over distinct primes.
For every prime $\ell$, put
\[
 E_\ell=\sum_{r<p,\,s<S}\min(\lfloor r/\ell\rfloor,S-1-s)
       =\sum_{k\ge1}(S-k)_+(p-k\ell)_+,
\]
where $x_+=\max(x,0)$. Put $\alpha=p/S$.
The corresponding integral is $\int_0^\infty(S-t)_+(p-\ell t)_+\,dt=Sp^2w_\alpha(\ell)/2$, where
\[
 w_\alpha(t)=\frac1{\max(\alpha,t)}
       -\frac{\min(\alpha,t)}{3\max(\alpha,t)^2}.
\]
Write $T_\ell=\sum_{r<p}\lfloor r/\ell\rfloor$, $V_\ell=\sum_{r<p}v_\ell(r!)$ and
$L=\sum_{r<p}\log r!$. Lemma~\ref{rows} gives $E_\ell\ge(S\ell w_\alpha(\ell)-1)T_\ell$.
Together with Legendre's bounds
\[
 (\ell-1)V_\ell\le\ell T_\ell,\qquad
 T_\ell\le V_\ell\le\frac{p^2}{2(\ell-1)},
\]
and $0\le\ell w_\alpha(\ell)\le1$, this yields
\[
 E_\ell\ge(S-1)V_\ell-\frac{Sp^2}2\left(\frac1{\ell-1}-w_\alpha(\ell)\right).
\]
Since $\sum_\ell V_\ell\log\ell=L$, summing over primes gives
\[
 \sum_\ell E_\ell\log\ell\ge(S-1)L-\frac{Sp^2}2C_p(\alpha),\qquad
 C_p(\alpha)=\sum_{\substack{\ell\le p\\\ell\ \mathrm{prime}}}\log\ell\left(\frac1{\ell-1}-w_\alpha(\ell)\right).
\]
All prime sums here are restricted to $\ell\le p$.
Lemma~\ref{rows} also gives $C_p(\alpha)<3.5$, since $40\le\alpha<40.001$ and $\log\alpha<3.69$.
Removing the non-inert primes costs at most $5.2Sp^2/2$, by~\eqref{scarce} and
$E_\ell\le ST_\ell\le Sp^2/(2\ell)$.
Using $L\ge\frac{p^2}2(\log p-\frac32)-p\log p$, we obtain
\begin{equation}\label{lower}
 \log D\ge Sp^2(\log p-10.21),
\end{equation}
where $2\log p/p+\log p/S<0.01$ covers the discrete error.

Comparing the two bounds for $\log D$, and using~\eqref{height} and $p>q^3/(5\log q)$, gives
\[
 \tfrac{47}{40}\log q-\tfrac{19}{40}\log(5\log q)<9.52.
\]
The left side increases with $q$. At $q=17000$, it exceeds
\[
 \frac{47}{40}(9.74)-\frac{19}{40}(3.89)=9.59675>9.52.
\]
Thus $q<17000$.

It remains to use~\eqref{inert} below $17000$. For $q>124$, the tests at $3$ and $5$ give
$q=120k+a$ with $a=43$ or $67$ and $0\le k<142$.
The tests at $7,11,13,17,31$ give
\[
\begin{array}{c|c|c|l}
 a & k\pmod7 & k\pmod{11} & k\\ \hline
 43 & 0,1,3 & 1,5,6,7,9 & 1,49,122\\
 67 & 0,4,5 & 0,3,7,8,9 & 117
\end{array}
\]
To obtain the last column, combine residues $u$ modulo $7$ and $v$ modulo $11$ by
$k\equiv22u-21v\pmod{77}$, then apply the tests at $13,17,31$.
This leaves $q=163,5923,14683,14107$.
For the last three, respectively, $-q\equiv14^2,7^2,4^2\pmod{29}$,
contradicting~\eqref{inert}.
For $8<q\le124$, the applicable tests at $3$ and $5$ leave $11,19,43,67$.
The fundamental discriminants with $q\le8$ are $-3,-4,-7,-8$.
For each of the nine surviving discriminants, Gauss reduction~\cite[arts.~171--175]{Gauss}
$|b|\le a\le c$, $3a^2\le q$ leaves only the principal form.
\end{proof}

\begin{lemma}[Row weights]\label{rows}
Let $p,S$ be positive integers with $\alpha=p/S\ge4$, and define
$E_\ell,T_\ell,w_\alpha$ and $C_p(\alpha)$ as above. For every prime $\ell$,
\[
 E_\ell\ge\bigl(S\ell w_\alpha(\ell)-1\bigr)T_\ell,
 \qquad
 C_p(\alpha)\le\frac{\log^2\alpha-\log\alpha+11}{6}.
\]
\end{lemma}

\begin{proof}
Put $u=p/\ell$, $H=\sum_{k\ge1}(u-k)_+$ and
$F=\sum_{k\ge1}(S-k)_+(u-k)_+$. Then $T_\ell=\ell H$ and $E_\ell=\ell F$.
The claim is immediate if $H=0$.
For $u\ge S$, direct summation and $H\le u^2/2$ give
\[
 F=\frac{S(S-1)}2\left(u-\frac{S+1}3\right),\qquad
 \frac FH\ge\frac{S^2}u-\frac{S^3}{3u^2}-1,
\]
the difference in the latter inequality being at least $1-S/u+S/(3u^2)\ge0$.
For $u\le S$, write $J=\sum_{k\ge1}k(u-k)_+$ and $n=\lfloor u\rfloor$.
The bound $F/H\ge S-u/3-1$ follows from
\[
 F=SH-J,\qquad
 (u+2)H-3J=n\left((u-n)^2+\frac{n-1}2\right)\ge0.
\]
For the second, put $f(t)=\log t\bigl((t-1)^{-1}-w_\alpha(t)\bigr)$.
This is nonnegative and convex for $t\ge2$.
On $t\le\alpha$, write
$f(t)=\log t/(t-1)-\log t/\alpha+t\log t/(3\alpha^2)$ and use
\[
 \frac{\log t}{t-1}=\int_0^1\frac{dv}{1+v(t-1)}.
\]
On $t\ge\alpha$, write $f(t)=\log t/[t(t-1)]+(\alpha/3)\log t/t^2$.
The first term is a product of positive decreasing convex functions, and the second has
nonnegative second derivative. The values and first derivatives agree at $t=\alpha$.
For $c=6j+5$ or $6j+7$, integrate $f(c-u)+f(c+u)\ge2f(c)$ over $0\le u\le3$.
Every prime beyond $3$ has one of these forms, so
\[
 C_p(\alpha)\le f(2)+f(3)+\frac16\int_2^\infty f(t)\,dt
                         +\frac16\int_4^\infty f(t)\,dt.
\]
For $2\le b\le\alpha$, direct integration gives
\[
 \begin{aligned}
 \int_b^\infty f(t)\,dt
 &=\frac{\log^2\alpha-\log^2b-\log\alpha}{2}+\frac54
   +\int_b^\infty\frac{\log t}{t(t-1)}\,dt\\
 &\quad+\frac b\alpha(\log b-1)+\frac{b^2}{12\alpha^2}(1-2\log b).
 \end{aligned}
\]
The remaining integral is at most
\[
 \int_b^\infty\frac{\log t}{(t-1)^2}\,dt
 =\frac{\log b}{b-1}+\log\frac b{b-1}.
\]
Substituting $b=2,4$ and collecting terms yields
\[
 C_p(\alpha)\le\frac{\log^2\alpha-\log\alpha}{6}
 +\frac{16}9\log2+\frac13\log3+\frac5{12}(1-(\log2)^2)
 -\frac1\alpha+\frac2{\alpha^2}.
\]
Here we used $0.69<\log2<0.70$ and $1<\log3<1.1$ to bound the coefficients of
$\alpha^{-1}$ and $\alpha^{-2}$ by $-1$ and $2$.
The same bounds make the constant terms less than $11/6$, while
$-1/\alpha+2/\alpha^2\le0$.
\end{proof}

\medskip
Gauss writes forms as $ax^2+2bxy+cy^2$~\cite[art.~153]{Gauss}.
Our form $(a,b,c)$ corresponds to his improperly primitive form $(2a,b,2c)$ when $b$ is odd,
and to his properly primitive form $(a,b/2,c)$ when $b$ is even~\cite[art.~226]{Gauss}.
Arithmetic in $\mathcal O$ is arithmetic on pairs of integers, with the displayed multiplication rule.
Modulo $p$, the inversion formula supplies everything needed for the polynomial root count.
The weighted prime sum uses only integration and finite addition.
Determinant expansion and row operations appear in Laplace~\cite[\S IV]{Laplace}.
Legendre's factorial formula appears in~\cite[Introduction, art.~XVI]{Legendre}.

When the class number is one, the principal form represents every non-inert prime.
Lattice counting and factorial valuations then force a large inert prime divisor of the central binomial coefficient,
by a comparison similar to Chebyshev's later work~\cite{Chebyshev}.

{\small
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\bibitem{Laplace} P.-S.~Laplace, \emph{Recherches sur le calcul int\'egral et sur le syst\`eme du monde},
M\'em.\ Acad.\ Sci.\ Paris, ann\'ee 1772, seconde partie (1776), 267--376.
\bibitem{Chebyshev} P.~L.~Chebyshev, \emph{M\'emoire sur les nombres premiers}, J.\ Math.\ Pures
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\end{thebibliography}
}
\end{document}