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\title{Extra Credit Problem}
\author{William Slatton}
\begin{document}
\maketitle
\section{Problem}
\[let \, y=(Ax^2+Bx+C)e^{Dx}\]
\[find \, A,B,C,D \in \mathbb{Z} \, such\, that \, y, \frac{dy}{dx}, \frac{dy^2}{dx^2} \, factor\, over \, \mathbb{Q}\]
\section{Solution}
\begin{proof}
\[\frac{dy}{dx}=(ADx^2+(BD+2A)x+(CD+B))e^{Dx}\]
\[\frac{dy^2}{dx^2}=(AD^2x^2+(BD^2+4AD)x+(CD^2+2A))e^{DX}\]
\[in \, order \, for \, these \, to \, factor, \, the \, quadratic\, elements\, must\, have\, roots\, in \, \mathbb{Q}\]
\[\therefore \, their\, respective\, discriminants\, must\, be\, perfect \, squares\]
\[\Delta \, of\, y=B^2-4AC=n^2_0 \, where\, n_0 \in \mathbb{Z}\]
\[\Delta \, of\, \frac{dy}{dx}=B^2D^2-4ACD^2+4A^2=(Dn_0)^2+4A^2=n^2_1 \, where\, n_1 \in \mathbb{Z}\]
\[\Delta \, of \, \frac{dy^2}{dx^2}=B^2D^4-4ACD^4+8A^2D^2=n^2_2 \, where\, n_2 \in \mathbb{Z}\]
\[(2A)^2+(Dn_0)^2=n^2_1\]
\[D^2((Dn_0)^2+8A^2)=n^2_2\]
\[(2A)^2+n^2_1=u^2 \, where\, u=n_2/D, \, which\, must\, necessarily\, be \, in \, \mathbb{Z}\]
\[no\, solution\, for\, A,B,C,D \neq 0\, by\, Fermat's\, Right\, Triangle\, Theorem\]
\end{proof}
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