Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$

Using congruences, show that the following is always an integer for every integer value of $n$: $$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$

Solution 1:

Lets show that $P(n)=3n^5+5n^3+7n$ is divisible by 15 for every $n$. To do this, we will show that it is divisible by $3$ and $5$ for every $n$.

Recall that for a prime $p$, $x^p\equiv x \pmod{p}$. (Fermat's Little Theorem) Then, looking modulo 5 we see that $$P(n)\equiv 3n^5+7n\equiv 3n+7n=10n\equiv 0.$$ Now looking modulo 3 we see that $$P(n)\equiv 5n^3+7n\equiv 5n+7n=12n\equiv 0.$$ Thus $P(n)$ is divisible by 15 for every $n$ as desired.

Solution 2:

HINT $\displaystyle\rm\quad \frac{n^5}5\: +\: \frac{n^3}3\: +\: \frac{7\:n}{15}\ =\ \frac{n^5-n}5\: +\: \frac{n^3-n}3\: +\: n\ \in \mathbb Z\ $ by Fermat's Little Theorem.

Solution 3:

Taking the lcm we have $\displaystyle \frac{1}{15} \cdot \Bigl[ 3n^{5}+5n^{3} + 7n\Bigr]$ Now show that the quantity $3n^{5} + 5n^{3}+7n$ is always divisible by $15$. Induction may be useful.

Clearly for $n=1$, $3+5+7=15$ is divisible by $15$. Assume that it is true for $n=k$. That is assume that $3k^{5}+5k^{3}+7k$ is divisible by $15$. Use this to show that the quantity

\begin{align*} 3(k+1)^{5}+5(k+1)^{3}+7(k+1) &= 3 \Bigl[k^{5} + {5 \choose 1}k^{4} + \cdots +1\Bigr] + 5(k+1)^{3}+ 7(k+1) \end{align*}

is divisible by $15$.