Show that
$$ \frac{12n^{13} + 23n}{35} $$
is an integer where $n$ is a natural number.
We follow the same strategy as the previous exercise.
We reformulate the requirement for the fraction to be an integer as a congruence.
$$ 12n^{13} + 23n \equiv 0 \pmod {35} $$
Following the hint we consider modulo 7 and modulo 5.
Using $n^7 \equiv n \pmod 7$, we have
$$ n^{13} \equiv n^7 \times n^6 \equiv n \times n^6 \times n^7 \equiv n \pmod 7 $$
And so
$$ 12n^{13} + 23n \equiv 12n + 23n \equiv 35n \equiv 0 \pmod {7} $$
Using $n^5 \equiv n \pmod 5$, we have
$$ n^{13} \equiv (n^5)^2 \times n^3 \equiv n^2 \times n^3 \equiv n^5 \equiv n \pmod 5 $$
And so
$$ 12n^{13} + 23n \equiv 12n + 23n \equiv 35n \equiv 0 \pmod {5} $$
We have
$$ 12n^{13} + 23n \equiv 0 \pmod {7} $$
$$ 12n^{13} + 23n \equiv 0 \pmod {5} $$
And since $\gcd(5,7)=1$, we conclude
$$ 12n^{13} + 23n \equiv 0 \pmod {35} $$
This is equivalent to the following fraction being an integer, for natural $n$
$$ \frac{12n^{13} + 23n}{35} $$