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\begin{document}
  \pagenumbering{gobble}
  \section{Q4}
We aim to prove that for every $x \in \mathbb{Z}$, the following is true:
\begin{equation}\tag{$\ast$}
p(x) \defeq \frac{1}{5}x^5 + \frac{1}{3}x^3 + \frac{7}{15}x \in \mathbb{Z}.
\end{equation}
First note that $p(0) = 0$ since:
\begin{align*}
p(0) &= \frac{1}{5}\cdot 0^5 + \frac{1}{3}\cdot 0^3 + \frac{7}{15}\cdot 0\\
&= 0 \in \mathbb{Z}.
\end{align*}
So $(\ast)$ is true for $x = 0$.\\
We now use induction to prove that $(\ast)$ is true for all $x \in \mathbb{Z}^+$. The statement holds for the base case $p(1)$:
\begin{align*}
p(x) &= \frac{1}{5}\cdot 1^5 + \frac{1}{3}\cdot 1^3 + \frac{7}{15}\cdot 1\\
&= \frac{1}{5} + \frac{1}{3} + \frac{7}{15}\\
&= 1 \in \mathbb{Z}.
\end{align*}
We now assume that $(\ast)$ is true for $x = k$, i.e. that $p(k) \in \mathbb{Z}$. To prove that $p(k+1) \in \mathbb{Z}$, we instead prove that $p(k+1) - p(k) = c$, for some $c \in \mathbb{Z}$. So:
\begin{align*}
p(k+1) - p(k) &= \frac{1}{5}(k+1)^5 + \frac {1}{3}(k+1)^3 + \frac {7}{15}(k+1) - \left(\frac{1}{5}k^5 + \frac{1}{3}k^3 + \frac{7}{15}k \right)\\
&=  \frac{(k+1)^5}{5} + \frac{(k+1)^3}{3} + \frac {7(k+1)}{15} - \frac{k^5}{5} - \frac{k^3}{3} - \frac{7k}{15}\\
&= \frac{(k+1)^5}{5} - \frac{k^5}{5} + \frac{(k+1)^3}{3} - \frac{k^3}{3} + \frac{7k}{15} - \frac{7k}{15} + \frac{1}{15}\\
&= \frac{1k + 5k + 10k^2 + 1-k^3 + 5k^4 + k^5 - k^5}{5}\\
&\quad + \frac{1 + 3k + 3k^2 + k^3 - k^3}{3} + \frac{7}{15}, \quad\text{by the binomial theorem}\\
&= \frac{1}{5} + \frac{5k}{5} + 2k^2 + 2k^3 + k^4 + \frac{1}{3} + k + k^2 + \frac{7}{15}\\
&= 2k + 3k^2 + 2k^3 + k^4 + \frac{1}{3} + \frac{1}{5} + \frac{7}{15}\\
&= 1 + 2k + 3k^2 + 2k^3 + k^4 \\
&\eqdef c,
\end{align*}
since products of integers are integers, and sums of integers are also integers, so $c \in \mathbb{Z}$.\\
So $p(k+1) - p(k) = c$ implies that $p(k+1) = c + p(k)$, and since $c, p(k) \in \mathbb{Z}$, we have that $p(k+1) \in \mathbb{Z}$, which proves that $(\ast)$ is true for all $x \in \mathbb{Z}^+$.\\
To prove this also works for negative integers, we consider instead:
$$p(-x) = - \frac{1}{5}x^5 - \frac{1}{3}x^3 - \frac{7}{15}x.$$
\end{document}