\documentclass{article}

\usepackage{amsmath, amssymb, amsthm}
\usepackage[margin=0.5in]{geometry}

\newcommand{\scs}[1]{\begin{scshape}#1\end{scshape}}
%\newcommand{\inv}{^{-1}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\RR}{\mathbb{R}}

\newtheorem{lemma}{Lemma}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newtheorem{problem}{}
\newtheorem*{challenge}{Challenge}
\newtheorem*{solution}{Solution}

\newcommand{\lem}[1]{\begin{lemma}#1\end{lemma}}
\newcommand{\prob}[1]{\begin{problem}#1\end{problem}} %\vfill}
\newcommand{\dfn}[1]{\begin{definition}#1\end{definition}}
\newcommand{\chal}[1]{\begin{challenge}#1\end{challenge}}
\newcommand{\sol}[1]{\begin{solution}#1\end{solution}}

\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\m}[1][n]{\left(\text{mod}\ #1\right)}

\begin{document} \pagestyle{empty} \begin{flushleft}

%\scs{Name(s):} 

\scs{Math 301 \hfill Proofs and disproofs \hfill \today}

\vspace{.1in}

Determine if each statement is true or false.
If it is true, prove it.
If it is false, give a disproof.

\prob{
There is an integer $n$ such that $n^2 \equiv -1 \m[13]$.
}
% TRUE, EXISTENCE PROOF

\sol{
The statement is true.
An example suffices to prove that such an integer exists: $ 2(13) = 26 = 25-(-1)$.
Hence $5^2 \equiv -1 \m[13]$.
}

\prob{
For all $a,b,c \in \ZZ$, if $a|bc$, then $a|b$ or $a|c$.
}
% FALSE, EXISTENCE PROOF

\sol{
The statement is false.
A counterexample suffices to prove the negation: there are $a,b,c \in \ZZ$ such that $a|bc$ but $a \nmid b$ and $a \nmid c$.
Let $a = 6$, $b=2$, and $c=3$.
We see that $6|6$ (so $a|bc$) but $6\nmid 2$ (so $a \nmid b$) and $6 \nmid 3$ (so $a \nmid c$). 
}

\prob{
There are integers $a$ and $b$ such that $12 a + 15 b = 2$.
}
% FALSE, UNIVERSAL PROOF

\sol{
The statement is false.
A disproof entails proving that for any integers $a$ and $b$ it follows that $12a + 15b \neq 2$.
We proceed by contradiction.
Suppose $a$ and $b$ are integers such that $12a + 15 b = 2$.
It follows that $3(4a+5b) = 2$ and thus that $3|2$.
This is a contradiction.
}


\prob{
There are integers $a$ and $b$ such that $11 a + 15 b = 1$.
}
% TRUE, EXISTENCE PROOF

\sol{
The statement is true.
An example suffices to prove it: $11(-4) + 15(3) = 1$.
}

\prob{
If $X \subseteq A \cup B$, then $X \subseteq A$ or $X \subseteq B$.
}
% FALSE, EXISTENCE PROOF

\sol{
The statement is false.
A counterexample suffices to prove the negation: there are sets $X$, $A$, and $B$ such that $X \subseteq A \cup B$ but $X \nsubseteq A$ and $X \nsubseteq B$.
Let $X = \{1,2\}$, $A = \{1\}$, and $B = \{2\}$.
Then $A \cup B = \{1,2\} \supseteq X$ but $X \nsubseteq A$ and $X \nsupseteq B$.
}

\chal{
There is an integer $n$ such that $n^2 \equiv -1 \m[3]$.
}
% FALSE, UNIVERSAL PROOF

\sol{
This statement is false.
A disproof entails proving that for any integer $n$ it follows that $n^2 \not\equiv -1 \m[3]$.
The proof I have in mind is direct and involves three cases: $n \equiv 0 \m[3]$, $n \equiv 1 \m[3]$, and $n \equiv 2 \m[3]$.
We can then determine that in the first case $n^2 \equiv 0 \m[3]$ and in the latter two cases $n^2 \equiv 1 \m[3]$.
In none of these cases is $n^2 \equiv -1 \m[3]$.
}

\end{flushleft} \end{document}