Problem 1620

1. Introduction to the problem:
We are given a polynomial of odd degree $2m+1$ over $\mathbb{Z}$:
$$f(x) = c_{2m+1}x^{2m+1} + \cdots + c_1 x + c_0$$
We need to show that this polynomial is irreducible if there exists a prime $p$ such that:
$$p \nmid c_{2m+1}, \quad p \mid c_{m+1}, c_{m+2}, \ldots, c_{2m}, \quad p^2 \mid c_0, c_1, \ldots, c_m, \quad \text{and} \quad p^3 \nmid c_0.$$

2. Using Newton polygons:
We will use the concept of Newton polygons in the field of $p$-adic numbers $\mathbb{Q}_p$. The $p$-adic valuation $v_p$ of a number is the exponent of the highest power of $p$ dividing that number.

3. Constructing the Newton polygon:
For the polynomial $f(x) = a_n x^n + \cdots + a_1 x + a_0$, we consider the set of points $(k, v_p(a_k))$. The Newton polygon is the lower convex hull of these points.

4. Analyzing the given conditions:
- $p \nmid c_{2m+1}$ implies $v_p(c_{2m+1}) = 0$.
- $p \mid c_{m+1}, c_{m+2}, \ldots, c_{2m}$ implies $v_p(c_{m+1}), v_p(c_{m+2}), \ldots, v_p(c_{2m}) \geq 1$.
- $p^2 \mid c_0, c_1, \ldots, c_m$ implies $v_p(c_0), v_p(c_1), \ldots, v_p(c_m) \geq 2$.
- $p^3 \nmid c_0$ implies $v_p(c_0) = 2$.

5. Constructing the Newton polygon:
The points we consider are:
$$(0, 2), (1, 2), \ldots, (m, 2), (m+1, 1), \ldots, (2m, 1), (2m+1, 0)$$
The Newton polygon will be a line segment from $(0, 2)$ to $(2m+1, 0)$.

6. Slope of the Newton polygon:
The slope of the line segment from $(0, 2)$ to $(2m+1, 0)$ is:
$$\text{slope} = \frac{0 - 2}{2m+1 - 0} = -\frac{2}{2m+1}$$

7. Roots of the polynomial:
According to the properties of Newton polygons, the roots of $f(x)$ in $\mathbb{Q}_p$ have valuations equal to the negative of the slope of the segments. Therefore, all roots of $f(x)$ have valuation $\frac{2}{2m+1}$.

8. Irreducibility argument:
Suppose $f(x) = g(x)h(x)$ where $g(x)$ and $h(x)$ are polynomials in $\mathbb{Q}_p[x]$. Then the valuations of the roots of $g(x)$ and $h(x)$ must sum to the valuations of the roots of $f(x)$. Since all roots of $f(x)$ have valuation $\frac{2}{2m+1}$, the degree of $g(x)$ must be a multiple of $2m+1$ to ensure that the valuation of the constant term of $g(x)$ is an integer. This implies that the only possible degrees for $g(x)$ are $0$ or $2m+1$, meaning $f(x)$ is irreducible.

$\blacksquare$