Integer Series Expansion

Question

For any two $p,q \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$, can one prove that

$a_n = \frac{p(-p)^n - q(p-2q)^n}{(p-q)}$

is an integer with recursion relation

$a_0 = 1,$ $a_n = a_{n-1}(p-2q)+2(-p)^n$?

Answer 1

$$\frac{p(-p)^n-q(p-2q)^n}{p-q}=\frac{p(-p)^n-q((p-q)-q)^n}{p-q}=\frac{p(-p)^n-q\sum_{i=0}^{n}\binom{n}{i}(p-q)^i(-q)^{n-i}}{p-q}$$$$=\frac{p(-p)^n-q(-q)^n}{p-q}-q\frac{\sum_{i=1}^{n}\binom{n}{i}(p-q)^i(-q)^{n-i}}{p-q}$$ The second term is obviously an integer number. The first term is equal to:

$$\frac{p(-p)^n-q(-q)^n}{p-q}=\frac{(-q)^{n+1}-(-p)^{n+1}}{p-q}$$ It is completely obvious that the nominator is a factor of $p-q$ (since $a^n-b^n=(a-b)\sum_{i=0}^{n-1}{a^{n-1-i}}{b^i}$). So $a_n$ is an integer number. $$a_{n-1}(p-2q)+2(-p)^n=\frac{p(-p)^{n-1}-q(p-2q)^{n-1}}{p-q}(p-2q)+2(-p)^n=\frac{p(p-2q)(-p)^{n-1}-q(p-2q)^n}{p-q}+2(-p)^n=\frac{p(p-2q)(-p)^{n-1}-q(p-2q)^n-2p(p-q)(-p)^{n-1}}{p-q}$$$$=\frac{p(-p)^{n-1}(p-2q-2p+2q)-q(p-2q)^n}{p-q}=\frac{p(-p)^n-q(p-2q)^n}{p-q}=a_n$$