Exercise 2.1.27 ($\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is an integer)

Question

Prove that $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is an integer for every integer $n$.

Richard Ganaye · Accepted Answer

Let 
$$N = \frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n = \frac{3n^5 + 5n^3 + 7n}{15}.$$
To prove that $N$ is an integer, we prove $15 \mid 3n^5 + 5n^3 + 7n$.

First, reducing modulo $3$,
\begin{align*}
3n^5 + 5n^3 + 7n &\equiv -(n^3 -n)\
&\equiv 0 \pmod 3,
\end{align*}
by Fermat's theorem.
Next, reducing modulo $5$,
\begin{align*}
3n^5 + 5n^3 + 7n &\equiv 3(n^5 - n)\
&\equiv 0 \pmod 5,
\end{align*}
by the same theorem.

Therefore, $3 \mid 3n^5 + 5n^3 + 7n$, and $5 \mid 3n^5 + 5n^3 + 7n$, where $3 \wedge 5 = 1$, thus $15 \mid 3n^5 + 5n^3 + 7n$.

This proves that $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is an integer for every integer $n$.

Exercise 2.1.27 ($\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is an integer)

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Exercise 2.1.27 ($\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is an integer)

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