We rewrite as $\frac{3a^5+5a^3+7a}{15}$. We can now test residues $\pmod {15}$. For example, plugging in $a=1$ gives an integer result, plugging in $a=2$ gives an integer result, and so on. For simplicity, just write as $3a^5+5a^3+7a \equiv 0 \pmod {15}$ and compute the values of each term $\pmod {15}$. Using these techniques the work isn't too bad and we find that every residue gives that $3a^5+5a^3+7a \equiv 0 \pmod {15}$, implying that for any integer $a$, the number $\frac{a^5}{5}+\frac{a^3}{3}+\frac{7a}{15}$ is an integer. $\blacksquare$

$3a^5+5a^3+7a\equiv 0+5a+7a\equiv 3(4a)\equiv 0$

$3a^5+5a^3+7a\equiv 3a+0+7a\equiv 5(2a)\equiv 0$