Problem 1469

1. Lemma 1: If $p$ is prime, then $\binom{p}{x}$ is always divisible by $p$ for $1 \le x \le p-1$.

Proof of Lemma 1:
$$\binom{p}{x} = \frac{p!}{x!(p-x)!}$$
Notice that the denominator $x!(p-x)!$ is the product of numbers less than $p$, so it can't possibly be divisible by $p$. Since the numerator $p!$ is divisible by $p$, and the expression is obviously an integer, we can conclude that $p \mid \binom{p}{x}$.

$\blacksquare$

2. Only if direction: Assume $p$ is prime. We need to show that all coefficients of the polynomial $F(x) = (x-a)^p - (x^p - a)$ are divisible by $p$.

- For $p = 2$:
$$(x-a)^2 - (x^2 - a) = x^2 - 2ax + a^2 - x^2 + a = -2ax + a^2 + a$$
Notice that:
$$-2ax + a^2 + a \equiv a^2 + a \equiv a(a+1) \equiv 0 \pmod{2}$$
Hence, all coefficients are divisible by $2$.

- For $p$ being an odd prime:
Using the binomial expansion:
$$(x-a)^p = x^p - \binom{p}{1}x^{p-1}a + \binom{p}{2}x^{p-2}a^2 - \cdots + (-a)^p$$
Therefore:
$$(x-a)^p - (x^p - a) = x^p - \binom{p}{1}x^{p-1}a + \binom{p}{2}x^{p-2}a^2 - \cdots + (-a)^p - x^p + a$$
Simplifying, we get:
$$-\binom{p}{1}x^{p-1}a + \binom{p}{2}x^{p-2}a^2 - \cdots + (-a)^p + a$$
By Lemma 1, each $\binom{p}{i}$ for $1 \le i \le p-1$ is divisible by $p$. Hence, all coefficients are divisible by $p$.

3. If direction: Assume all coefficients of $F(x) = (x-a)^p - (x^p - a)$ are divisible by $p$. We need to show that $p$ is prime.

- Suppose $p$ is not prime. Let $d$ be the largest prime factor of $p$. Consider:
$$\binom{p}{d} = \frac{p(p-1)(p-2)\cdots (p-d+1)}{d!}$$
It is clear that exactly one of the terms in the numerator can be divisible by $d$, so that must be $p$. Since the denominator $d!$ must have one factor of $d$ (since $d$ is prime), $\binom{p}{d}$ has one less factor of $d$ than $p$. Therefore, $\binom{p}{d}$ can't possibly be divisible by $p$.

- Since $\binom{p}{d}$ is one of the coefficients (possibly multiplied by $\pm 1$), not all of the coefficients are divisible by $p$ if $p$ is not prime. This contradicts our assumption that all coefficients are divisible by $p$.

Hence, $p$ must be prime.

$\blacksquare$