Problem 1724

To find all integers $a, b, m, n$ with $m > n > 1$ such that the polynomial $f(X) = X^n + aX + b$ divides the polynomial $g(X) = X^m + aX + b$, we start by expressing $m$ in terms of $n$. Let $m = n + k$ for some $k \in \mathbb{N}$.

1. Divisibility Condition:
$$f(X) \mid g(X) \implies f(X) \mid g(X) - f(X) = X^m - X^n = X^n(X^k - 1)$$
This implies that $f(X)$ must divide $X^n(X^k - 1)$.

2. Roots Analysis:
Since the roots of $X^n(X^k - 1)$ are all of magnitude $0$ or $1$, the roots of $f(X) = X^n + aX + b$ must also be of magnitude $0$ or $1$. Therefore, $b$ must be $0$ or $\pm 1$.

### Case I: $b = 0$
$$f(X) = X^n + aX$$
$$f(X) \mid X^n(X^k - 1) \implies X^{n-1} + a \mid X^{n-1}(X^k - 1)$$
The roots of $X^{n-1}(X^k - 1)$ are all of magnitude $0$ or $1$, so the roots of $X^{n-1} + a$ must also be of magnitude $0$ or $1$. Thus, $a = 0$ or $\pm 1$.

- **Subcase $a = 0$**:
$$f(X) = X^n \implies X^n \mid X^m \text{ for all } m > n > 1$$
Therefore, $(a, b, m, n) = (0, 0, m, n)$ is a solution for all $m > n > 1$.

- **Subcase $a = -1$**:
$$f(X) = X^{n-1} - 1 \implies X^{n-1} - 1 \mid X^k - 1$$
This holds if the $(n-1)$-th roots of unity are a subset of the $k$-th roots of unity. Therefore, $n-1 \mid k$, i.e., $k = t(n-1)$ for some $t \in \mathbb{N}$. Thus,
$$m = n + k = n + t(n-1) = (t+1)n - t$$
Therefore, $(a, b, m, n) = (-1, 0, (t+1)n - t, n)$ is a solution for all $n > 1$ and $t \in \mathbb{N}$.

- **Subcase $a = 1$**:
$$f(X) = X^{n-1} + 1 \implies X^{n-1} + 1 \mid X^k - 1$$
The roots of $X^{n-1} + 1$ are the $2(n-1)$-th roots of unity that are not $(n-1)$-th roots of unity. Therefore, $2(n-1) \mid k$, i.e., $k = 2t(n-1)$ for some $t \in \mathbb{N}$. Thus,
$$m = n + k = n + 2t(n-1) = (2t+1)n - 2t$$
Therefore, $(a, b, m, n) = (1, 0, (2t+1)n - 2t, n)$ is a solution for all $n > 1$ and $t \in \mathbb{N}$.

### Case II: $b = \pm 1$
$$f(X) = X^n + aX \pm 1$$
$$f(X) \mid X^n(X^k - 1) \implies f(X) \mid X^k - 1$$
Since $0$ is not a root of $X^n + aX \pm 1$, we consider the roots of $X^n + aX \pm 1$.

- **Subcase $n > 2$**:
By Vieta's formulas, the sum of the roots is zero, and the product of the roots is $\pm 1$. Since the polynomial has real coefficients, if $r_i$ is a root, then $\overline{r_i}$ must also be a root. For $r_i$ being $k$-th roots of unity, $\overline{r_i} = \frac{1}{r_i}$. Thus,
$$\sum r_i = \sum \frac{1}{r_i} \implies \pm a = 0 \implies a = 0$$

- **Subcase $a = 0, b = -1$**:
$$f(X) = X^n - 1 \implies X^n - 1 \mid X^k - 1$$
This holds if the $n$-th roots of unity are a subset of the $k$-th roots of unity. Therefore, $n \mid k$, i.e., $k = tn$ for some $t \in \mathbb{N}$. Thus,
$$m = n + k = n + tn = (t+1)n$$
Therefore, $(a, b, m, n) = (0, -1, (t+1)n, n)$ is a solution for all $n > 1$ and $t \in \mathbb{N}$.

- **Subcase $a = 0, b = 1$**:
$$f(X) = X^n + 1 \implies X^n + 1 \mid X^k - 1$$
The roots of $X^n + 1$ are the $2n$-th roots of unity that are not $n$-th roots of unity. Therefore, $2n \mid k$, i.e., $k = 2tn$ for some $t \in \mathbb{N}$. Thus,
$$m = n + k = n + 2tn = (2t+1)n$$
Therefore, $(a, b, m, n) = (0, 1, (2t+1)n, n)$ is a solution for all $n > 1$ and $t \in \mathbb{N}$.

- **Subcase $n = 2$**:
$$f(X) = X^2 + ax \pm 1 \mid X^k - 1$$
The roots of $X^2 + ax \pm 1$ must be $k$-th roots of unity. If the roots are $\pm 1$, then $f(X) = X^2 - 1$, which has already been covered. Otherwise, the roots are of the form $\text{cis}(\pm \theta)$.

- **Subcase $a = -1, b = -1$**:
$$f(X) = X^2 - X + 1 \implies X^2 - X + 1 \mid X^k - 1$$
The roots are $\text{cis}(\pm \frac{\pi}{3})$, which must be $k$-th roots of unity. Therefore, $6 \mid k$, i.e., $k = 6t$ for some $t \in \mathbb{N}$. Thus,
$$m = n + k = 2 + 6t = 2 + 6t$$
Therefore, $(a, b, m, n) = (-1, -1, 2 + 6t, 2)$ is a solution for all $t \in \mathbb{N}$.

- **Subcase $a = 1, b = -1$**:
$$f(X) = X^2 + X + 1 \implies X^2 + X + 1 \mid X^k - 1$$
The roots are $\text{cis}(\pm \frac{\pi}{3})$, which must be $k$-th roots of unity. Therefore, $3 \mid k$, i.e., $k = 3t$ for some $t \in \mathbb{N}$. Thus,
$$m = n + k = 2 + 3t = 2 + 3t$$
Therefore, $(a, b, m, n) = (1, -1, 2 + 3t, 2)$ is a solution for all $t \in \mathbb{N}$.

### Conclusion:
The solutions are:
$$(0, 0, m, n), \quad (-1, 0, (t+1)n - t, n), \quad (1, 0, (2t+1)n - 2t, n), \quad (0, -1, (t+1)n, n), \quad (0, 1, (2t+1)n, n), \quad (-1, -1, n + 6t, n), \quad (1, -1, n + 3t, n)$$
where $m, n, t \in \mathbb{N}$ such that $m > n > 1$.

The final answer is: