Let $ m = n + k$ for some $ k \in \mathbb{N}$. $ f(x)$ divides $ g(x)$ iff $ f(x)$ divides $ g(x) - f(x) = x^m - x^n = x^n(x^k - 1)$. 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 $ 0,1$. Thus, $ b = 0$ OR $ b = \pm 1$. Case I: $ b = 0$: $ x^n + ax | x^n(x^k - 1) \Longleftrightarrow x^{n - 1} + a | x^{n - 1}(x^k - 1)$ Since the roots of $ x^{n - 1}(x^k - 1)$ are all of magnitude $ 0$ or $ 1$, the roots of $ x^{n - 1} + a$ must also be of magnitude $ 0,1$. Thus, $ a = 0$ OR $ a = \pm 1$. $ (a,b) = (0,0)$ yields: $ x^n|x^m$ which is true for all integers $ m > n > 1$. Thus, $ (a,b,m,n) = (0,0,m,n)$ is a solution for all positive integers $ m > n > 1$. $ (a,b) = ( - 1,0)$ yields: $ x^{n - 1} - 1|x^k - 1$. $ x^{n - 1} - 1|x^k - 1$ iff the $ (n - 1)$th roots of unity are a subset of the $ k$th roots of unity. Therefore, $ n - 1|m - n \Leftrightarrow m - n = t(n - 1) \Leftrightarrow m = (t + 1)n - t$ for any $ t \in \mathbb{N}$. Thus, $ (a,b,m,n) = ( - 1,0,(t + 1)n - t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$. $ (a,b) = (1,0)$ yields: $ x^{n - 1} + 1|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. $ x^{n - 1} + 1|x^k - 1$ iff the $ 2(n - 1)$th roots of unity are a subset of the $ k$th roots of unity. Therefore, $ 2(n - 1)|m - n \Leftrightarrow m - n = 2t(n - 1) \Leftrightarrow m = (2t + 1)n - 2t$ for any $ t \in \mathbb{N}$. Thus, $ (a,b,m,n) = (1,0,(2t + 1)n - 2t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$. Case II: $ b = \pm1$: $ x^n + ax\pm1 | x^n(x^k - 1) \Longleftrightarrow x^n + ax\pm1 | x^k - 1$ (Since $ 0$ is not a root of $ x^n + ax\pm1$). Let the roots of $ x^n + ax \pm 1$ be $ r_i$ for $ i \in \overline{1,n}$. If $ n > 2$, by Vieta's: $ \sum r_i = 0$, $ \prod r_i = \pm1$, and thus, $ \sum \frac {1}{r_i} = \sum \frac {\pm r_1 r_2 \cdots r_{n - 1} r_n}{r_i} = \pm a$ However, since $ x^n + ax \pm 1$ has all real coefficients, if $ r_i$ is a root, then $ \overline{r_i}$ must also be a root. Note that since all the $ r_i$ are $ k$th roots of unity: $ \overline{r_i} = \frac {1}{r_i}$. Thus, $ r_i + \overline{r_i} = \frac {1}{r_i} + \frac {1}{\overline{r_i}}$ for all non-real $ k$th roots of unity $ r_i$. The other two possible $ k$th roots of unity are, $ \pm1$ which satisfy $ r_i = \frac {1}{r_i}$. Therefore, if $ n > 2$, then $ \sum \frac {1}{r_i} = \sum r_i \Longleftrightarrow \pm a = 0 \Rightarrow a = 0$. $ (a,b) = (0, - 1)$ yields: $ x^n - 1 | x^k - 1$. $ x^n - 1|x^k - 1$ iff the $ n$th roots of unity are a subset of the $ k$th roots of unity. Therefore, $ n|m - n \Leftrightarrow m - n = tn \Leftrightarrow m = (t + 1)n$ for any $ t \in \mathbb{N}$. Thus, $ (a,b,m,n) = (0, - 1,(t + 1)n,n)$ for all positive integers $ n,t$ such that $ n > 1$. $ (a,b) = (0,1)$ yields: $ x^n + 1 | x^k - 1$. The roots of $ x^n + 1$ are the $ 2n$th roots of unity that are not $ n$th roots of unity. $ x^n + 1|x^k - 1$ iff the $ 2n$th roots of unity are a subset of the $ k$th roots of unity. Therefore, $ 2n|m - n \Leftrightarrow m - n = 2tn \Leftrightarrow m = (2t + 1)n$ for any $ t \in \mathbb{N}$. Thus, $ (a,b,m,n) = (0,1,(2t + 1)n,n)$ for all positive integers $ n,t$ such that $ n > 1$. $ n = 2$ yields: $ x^2 + ax\pm1 | x^k - 1$. Since the roots of $ x^2 + ax\pm1$ are $ k$th roots of unity, either the roots are $ \pm 1$ (in which case $ f(x) = x^2 - 1$ which has already been covered above with $ (a,b) = (0, - 1)$), or in the form $ cis\left(\pm\theta\right)$. Therefore, $ x^2 + ax\pm1 = (x - cis(\theta))(x - cis( - \theta)) = x^2 - 2\cos(\theta)x + 1$. Thus, $ b = 1$ and $ a = - 2\cos(\theta)$ must be a non-zero integer, the only possibilities are $ \pm1$ and $ \pm2$. If $ a = \pm2$, then $ f(x)$ has a double root, which is a contradiction since $ x^k - 1$ does not have a double root. $ (a,b) = ( - 1, - 1)$ yields $ x^2 - x + 1|x^k - 1$. The roots of $ x^2 - x + 1$ are $ cis\left(\pm\frac {\pi}{6}\right)$ which must be $ k$th roots of unity. Therefore, $ 6|m - n \Leftrightarrow m - n = 6t \Leftrightarrow m = n + 6t$. Thus, $ (a,b,m,n) = ( - 1, - 1,n + 6t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$. $ (a,b) = (1, - 1)$ yields $ x^2 + x + 1|x^k - 1$. The roots of $ x^2 - x + 1$ are $ cis\left(\pm\frac {\pi}{3}\right)$ which must be $ k$th roots of unity. Therefore, $ 3|m - n \Leftrightarrow m - n = 3t \Leftrightarrow m = n + 3t$. Thus, $ (a,b,m,n) = ( - 1, - 1,n + 3t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$. Therefore, the solutions are: $ (0,0,m,n)$ $ ( - 1,0,(t + 1)n - t,n)$ $ (1,0,(2t + 1)n - 2t,n)$ $ (0, - 1,(t + 1)n,n)$ $ (0,1,(2t + 1)n,n)$ $ ( - 1, - 1,n + 6t,n)$ $ ( - 1, - 1,n + 3t,n)$ Where $ m,n,t \in \mathbb{N}$ such that $ m > n > 1$.