Integer solutions to $D(f)=0$ for $f(X)=X^n + aX+b$ By using the relation $D(f)=(-1)^{\binom{n}{2}}R(f,f')$ between the discriminant and the resultant, it is not hard to show that $D(f)=(-1)^{\binom{n}{2}}(n^nb^{n-1}-(n-1)^{n-1}(-a)^n)$ for $f(X)=X^n+aX+b$. Thus,
$D(f)=0$ holds iff $(\frac{-a}{n})^n=(\frac{b}{n-1})^{n-1}$.
How to find all integer solutions $a,b \in \mathbb{Z}$?
I'm tempted to require $\frac{-a}{n}=x^{n-1} \in \mathbb{Z}$ and $\frac{b}{n-1}=y^n \in \mathbb{Z}$ and find integer solutions $x,y \in \mathbb{Z}$ to the resulting equations, but I don't have a good explanation why this would give all solutions.
 A: In fact, some simple arithmetical arguments show that you get all solutions this way. Start from
$$
n^n b^{n-1} = (n-1)^{n-1}(-a)^n \label{1}\tag{1}
$$
We see from \eqref{1} that $n^n$ divides $(n-1)^{n-1}(-a)^n$ ; but since $n^n$ is coprime to $n-1$, $n^n$ must divide $(-a)^n$. Let $g=\gcd(a,n)$, and $n'=\frac{n}{g}$, $a'=\frac{a}{g}$. Then $(n')^n$ divides $(a')^n$, and hence $n'=\pm 1$ since $n'$ and $a'$ are coprime. In other words $n$ divides $a$, or
$A=\frac{-a}{n}$ is an integer. Similarly $B=\frac{b}{n-1}$ is an integer.
Then \eqref{1} becomes
$$
B^{n-1} = A^n \label{2}\tag{2}
$$
Let $p$ be any prime dividing $B$, let $\nu$ be the exponent of $p$ in $B$ so that $B=p^{\nu}C$ where $p$ does not divide $C$. Then the exponent of $p$ in $B^{n-1} = A^n$ is $(n-1)\nu$ ; but we know that this exponent must be a multiple of $n$. Since $n$ and $n-1$ are comprime, $\nu$ must be a multiple of $n$.
Since this holds for any prime divisor $p$ of $B$, $B$ is a perfect $n$-th power, $B=t^n$ for some integer $t$, and then $A=t^{n-1}$, which finishes the proof.
