\documentclass{amsart}
\usepackage{amsmath,amsthm,amssymb}

\thanks{The author was supported by NSF grant DMS-1148634}

\newtheorem{prop}{Proposition}[section]
\newtheorem{defn}[prop]{Definition}
\newtheorem{conj}[prop]{Conjecture}
\newtheorem{lemma}[prop]{Lemma}
\newtheorem{cor}[prop]{Corollary}
\newtheorem{theorem}[prop]{Theorem}
\newtheorem{example}[prop]{Example}
\newtheorem{remark}[prop]{Remark}
\newtheorem{sconj}[prop]{Super-Conjecture}

\begin{document}
{\centerline{June 16, 2016}}
\section{Super-Conjecture 1}


\author{Dennis Stanton}
\address{School of Mathematics, University of Minnesota, Minneapolis, MN 55455}
\email{stanton@math.umn.edu}


\begin{sconj} Let $p\ge 5$ be a prime,  and
$$
A=\sum_{n=0}^{p-1} \binom{2n}{n}.
$$
Then
$$
\begin{aligned}
A &\equiv 1 \mod p^2 {\text{   if $p\equiv 1 \mod 3$}},\\
A &\equiv -1 \mod p^2 {\text{ if $p\equiv 2 \mod 3$}}.
\end{aligned}
$$
\end{sconj}
\vskip10pt
\begin{proof} Already from the paper we know
$$
A=CT \frac{(1+x)^{2p}}{x^{p-1}(1+x+x^2)}=CT \frac{(1-x)(1+x)^{2p}}{x^{p-1}(1-x^3)}.
$$
So 
\begin{equation}
\label{sumforA}
A=\sum_{k\ge 0}\left( \binom{2p}{p-1-3k}-\binom{2p}{p-2-3k}\right)
\end{equation}
\vskip10pt
\underline{CASE 1} $p\equiv 1 \mod 3.$
In this case, the $k=(p-1)/3$ term in \eqref{sumforA} is 1, and all of the 
other terms are divisible by $p$,
$$
A=1+pB
$$
where 
$$
B=\frac{1}{p}\sum_{0\le k< (p-1)/3}\left( \binom{2p}{p-1-3k}-\binom{2p}{p-2-3k}\right).
$$
\vskip10pt
We'll show that the integer $B$ satsifies $B\equiv 0\mod p$, so that $A\equiv 1 \mod p^2.$
\vskip10pt
Since
$$
\begin{aligned}
\frac{1}{p} \binom{2p}{p-1-3k} &\equiv 2\frac{(p-1)!}{(p-1-3k)!(3k+1)!} \mod p,\\
\frac{1}{p} \binom{2p}{p-2-3k}& \equiv 2\frac{(p-1)!}{(p-2-3k)!(3k+2)!} \mod p
\end{aligned}
$$
we see that
$$
B\equiv 2\sum_{0\le k < (p-1)/3}\left( \frac{(p-1)!}{(p-1-3k)!(3k+1)!} -\frac{(p-1)!}{(p-2-3k)!(3k+2)!}
\right) \mod p
$$

However, by trisecting the binomial theorem this sum of integers is $0$
$$
\sum_{0\le k < (p-1)/3}\left( \frac{(p-1)!}{(p-1-3k)!(3k+1)!} -\frac{(p-1)!}{(p-2-3k)!(3k+2)!}
\right) =0,
$$
so
$$
B\equiv 0\mod p.
$$
\vskip10pt
\underline{CASE 2} $p\equiv 2 \mod 3.$ In this case the second term with $k=(p-2)/3$ gives $-1.$ 
We proceed as in CASE 1, the trisection identity we need is
$$
\sum_{0\le k \le (p-2)/3} \frac{(p-1)!}{(p-1-3k)!(3k+1)!} -
\sum_{0\le k \le (p-5)/3} \frac{(p-1)!}{(p-2-3k)!(3k+2)!} =0,
$$
\end{proof}

\section{Super-Conjecture 1''}
\begin{sconj} Let $p\ge 5$ be a prime,  $r$ be a positive integer, and
$$
A=\sum_{n=0}^{rp-1} \binom{2n}{n}.
$$
Then
$$
\begin{aligned}
A &\equiv \alpha_r \mod p^2 {\text{   if $p\equiv 1 \mod 3$}},\\
A &\equiv -\alpha_r \mod p^2 {\text{ if $p\equiv 2 \mod 3$}}.
\end{aligned}
$$
where
$$
\alpha_r=\sum_{n=0}^{r-1}\binom{2n}{n}.
$$
\end{sconj}

\begin{proof}(sketch, assuming a fact \eqref{isthistrue?} you probably know, if true) This time
$$
A=CT \frac{(1+x)^{2rp}}{x^{rp-1}(1+x+x^2)}=CT \frac{(1-x)(1+x)^{2rp}}{x^{rp-1}(1-x^3)}.
$$
So 
\begin{equation}
\label{sumforAr}
A=\sum_{k\ge 0}\left( \binom{2rp}{rp-1-3k}-\binom{2rp}{pr-2-3k}\right)
\end{equation}

As before, we separate the terms which have a $p$ for sure, namely those whose 
binomial denominator is not a multiple of $p$ from those whose binomial denominator 
is a multiple of $p$. The non-multiples of $p$ will sum to $0$ as before, and we are left 
with the extra terms.
\vskip10pt
\underline{CASE 1} $p\equiv 1 \mod 3.$ Here the terms with denominator parameter a 
multiple of $p$ are
$$
EXTRA=\sum_{j\ge 0} \left( \binom{2rp}{(r-1-3j)p}-\binom{2rp}{(r-2-3j)p}\right).
$$
\vskip10pt
Assuming
\begin{equation}
\label{isthistrue?}
\binom{Dp}{Ep}\equiv \binom{D}{E} \mod p^2
\end{equation}
then
$$
EXTRA\equiv \sum_{j\ge 0} \left( \binom{2r}{r-1-3j}-\binom{2r}{r-2-3j}\right)= \alpha_r \mod p^2
$$
by \eqref{sumforA}, which works for any positive integer $p$.

\vskip10pt 
\underline{CASE 2} $p\equiv 2 \mod 3.$ Here the terms with denominator parameter being a 
multiple of $p$ are
$$
EXTRA=\sum_{j\ge 0} \left( \binom{2rp}{(r-2-3j)p}-\binom{2rp}{(r-1-3j)p}\right).
$$
then
$$
EXTRA\equiv \sum_{j\ge 0} \left( \binom{2r}{r-2-3j}-\binom{2r}{r-1-3j}\right)= -\alpha_r \mod p^2
$$
again by \eqref{sumforA}. 


\end{proof}

\end{document}
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