Problem 27: Prove that \(\frac{1}{5}n^{5} + \frac{1}{3}n^{3} + \frac{7}{15}n\) is an integer for every integer \(n\).
Proof
Observe that
$$
\begin{align*}
\frac{1}{5}n^{5} + \frac{1}{3}n^{3} + \frac{7}{15}n = \frac{3n^5 + 5n^3 + 7n}{15}
\end{align*}
$$
To show that this is integer. This is equivalent to showing that
$$
\begin{align*}
3n^5 + 5n^3 + 7n
\end{align*}
$$
is divisible by \(15\). But we know that this is equivalent to showing that this expression is divisible by both \(3\) and \(5\). To show that it’s divisible by \(3\), observe that
$$
\begin{align*}
3n^5 + 5n^3 + 7n &\equiv 0 + 2n^3 + 1n \pmod{3} \\
&\equiv 2n^3 + n \pmod{3} \\
&\equiv n(2n^2 + 1) \pmod{3}
\end{align*}
$$
We can plug in \(n = 0,1,2\) to see that
$$
\begin{align*}
n&=0: 0(2\cdot (0)^2 + 1) \equiv 0 \pmod{3} \\
n&=1: 1(2\cdot (1)^2 + 1) = 3 \equiv 0 \pmod{3} \\
n&=2: 2(2\cdot (2)^2 + 1) = 18 \equiv 0 \pmod{3}
\end{align*}
$$
Therefore, \(3n^5 + 5n^3 + 7n\) is divisible by \(3\). A similar argument shows that it’s divisible by \(5\) which leads to the conclusion that it’s divisible by \(15\) as desired. \(\ \blacksquare\)