I'm trying to follow the main given answer here There is a positive integer $y$ such that for a polynomial with integer coefficients we have $f(y)$ as composite
What I don't understand is where in the binomial expansion at the top the other powers of x have gone to be replaced by a simple integer value?
Let us take an example of a quadratic. We can explicitly write everything in this case.
Set $$f(x) = a_2 x^2 + a_1 x + a_0$$ and consider $$f(x+kp) = a_2 (x^2 + 2xkp + (kp)^2) + a_1 (x+kp) + a_0$$ then moving terms around we find:
$$f(x+kp) = (a_2 x^2 + a_1 x +a_0) + (2a_2 xkp + a_2 k^2p^2 + a_1 kp)$$ $$= f(x) + (2a_2 xk + a_2 k^2p + a_1 k)p$$
Now we declare $K = (2a_2 xk + a_2 k^2p + a_1 k)$ and we have:
$$f(x+kp) = f(x) + Kp$$
You actually have $\alpha_n = \alpha_{n}(x)$ - it contains all powers of $x$. What matters for the proof is that all factors in the expansion after the first have at least $p$ (or a higher power of $p$), so you can factor it out and write all terms from the second as $\alpha_{n}(x) \cdot p$ ($\alpha_{n-1}(x) \cdot p$, ...), and so the sum of these terms can be divided by $p$.
