Theorem: The language $F = \setbuilder{a^i b^j}{i = kj \text{ for some } k}$ is not context-free.

---

Proof: Suppose $F$ were context-free, and let $p$ be the pumping length. Let $q$ be a number strictly greater than $p$, and let $z = a^{q^2} b^q$. Note $|z| \geq p$ and $z \in F$. Let $z = uvwxy$ with $|vwx| \leq p$ and $|vx| \geq 1$. Consider cases.

Suppose $v$ and $x$ consist entirely of $a$'s. That is, $v=a^n$ and $x = a^m$ for some $n, m$ with $1 \leq n+m \leq p$. Notice that $\#_a(u v^0 w x^0 y) = q^2 - (n+m)$. Now, we use the strict requirements on $q$ to note that:

$$q^2 - q < q^2 - (n+m) < q^2$$

with strict inequalities, so $q$ cannot possibly divide the middle quantity. Therefore, the number of $b$'s does not divide the number of $a$'s and $u v^0 w x^0 y \not\in F$.

Suppose $v$ and $x$ consist entirely of $b$'s. Then $u v^i w x^i y$ will only have a strictly increasing number of $b$'s as $i$ increases. So $i$ can certainly be picked such that $\#_b(u v^i w x^i y) > q^2$, then the the number of $b$'s does not divide the number of $a$'s, and $u v^i w x^i y \not\in F$.

In the case that $v$ or $x$ contains mixed characters, that is, some $a$'s and some $b$'s, then $u v^2 w x^2 y$ is not of the form $a^* b^*$, and not in $F$.

Finally, we consider the case where $v$ consists only of $a$'s and $x$ consists only of $b$'s. So $v = a^n$ and $x = b^m$ for some $n \geq 1$, $m \geq 1$, and $n+m \leq p$. Let $a(i) = \#_a(u v^i w x^i y)$ and $b(i) = \#_b(u v^i w x^i y)$. An $i$-pumped string is in $F$ if and only if $a(i)/b(i)$ is an integer. We have, for any positive $i$:

$$r_i = \dfrac{a(i)}{b(i)} = \dfrac{q^2 + (i-1)n}{q + (i-1)m}$$

Note that $r_1 = q$, $r_i$ is strictly decreasing, and converges to $n/m$ as $i \rightarrow \infty$. Since there are clearly only a finite number of integers (if any) between $q$ and $n/m$, there must be some $i$ such that $r_i$ is not an integer. So, for that choice of $i$, $u v^i w x^i y \not\in F$.

This contradicts the pumping lemma, so $F$ cannot be context-free.