The way I understand it, is that if $f(x)$ is an irreducible polynomial in $\mathbb{Q}[x]$ of degree at least 2, then a difference of distinct roots $a_i-a_j$ is never rational for any of the $a_1,\dots,a_n$ which are the roots of $f(x)$ in $\mathbb{C}$.
Why is this? If $a_i-a_j\in\mathbb{Q}$ for some distinct roots, what goes wrong? Would it follow somehow that $f(x)$ is reducible over $\mathbb{Q}$? Or perhaps there's a more direct explanation?
Best Answer
Let $a$ and $a+b$ be roots of $f$ with $b$ rational and nonzero. Then $$g(x):=f(x)-f(x+b)$$ satisfies $g\in\mathbb Q[x]$, $g\neq0$, $g(a)=0$ and $\deg g < \deg f$, contradiction.
[We assume $f\in\mathbb Q[x]$ and $f$ irreducible.]
EDIT 1. Variation:
Let $a$ and $a+b$ be roots of $f$ with $b$ rational and nonzero.
Then $f(x+b)=f(x)$ since $a$ is a root of $f(x+b)\in\mathbb Q[x]$, contradiction.
EDIT 2. I'll steal just one thing from Georges's great answer:
Let $K$ be a field, $d$ a nonzero element of $K$, and $f(X)\in K[X]$ an irreducible polynomial. Then $d$ is the difference of two roots of $f(X)$ if and only if $f(X+d)=f(X)$. If this is the case, the characteristic of $K$ divides the degree of $f(X)$.
The simplest occurrence of this phenomenon is for $K=\mathbb F_2$, $d=1$, $f(X)=X^2+X+1$.