A question about the divisibility of certain polynomial sequences.

$2n+1=(n+1)^2-(n)^2$ .
Therefore $(n+1)^2-n^2$ never divides $2$ for any integers.Can we make a similar statement for $(n+1)^x-n^x=a_n$
... And if we can, can we combine polynomials to give us a formula for integers that do not divide a certain set of integers.
I've tried binomial theorem but I have to deal with polynomial divisability , which is pretty much my original question.

Answer

For polynomials of the form $(n+1)^x-n^x$ you probably should use modular arithmetic.

For example, for $(n+1)^3-n^3$, we can check its divisibility by $3$. Clearly, if $n \equiv 0$ or $2$ then it is not divisible. Now we check the case $n \equiv 1$. $1^3 \equiv 1$ and $2^3 \equiv 2$ modulo $3$, so it would still not be divisible by $3$ in this case. ($2-1=1$)

As long as you don't have consecutive integers to a power having the same residue, it guarantees that it is not divisible.

Checking divisibility by $5$ for a power of $4$:
$$
0^4 = 0 \equiv 0 \\

1^4 = 1 \equiv 1 \\
2^4 = 16 \equiv 1 \\
3^4 = 81 \equiv 1 \\
4^4 = 256 \equiv 1 \\
$$

So for these values it clearly doesn't work, in fact most values of $n$ will end up being divisible by $5$.

The reason why all those ones appear is due to something called Fermat's Little Theorem. If you want to look into this problem better, begin with elementary number theory and modular arithmetic, then Fermat's Little Theorem, and if you want to go even further look up Carmichael numbers.