Proof that a given language is not context-free
Use the pumping lemma for CFLs. Let N be the pumping lemma's constant, and take $\sigma = a^{N^{3/2}}$, which certainly has $\lvert \sigma \rvert \ge N$. So we can write $\sigma = v w x y z$ with $\lvert w x y \rvert \le N$ and $w y \ne \epsilon$ such that for all $k \ge 0$ we have $v w^k x y^k z \in L$.

But here only lengths matter. Call $u = \lvert w y \rvert$, so that the length of the pumped string is:

$\begin{align}
\lvert v w^k x y^k z  \rvert = N^{3/2} + (k - 1) u
\end{align}$

where we know that $u \le N$.

Now look for $n$ such that $(n + 1)^{3/2} - n^{3/2} &gt; N$. By the binomial theorem:

$\begin{align}
(n + 1)^{3/2} - n^{3/2}
  &amp;=   n^{3/2} \left( (1 + 1/n)^{3/2} - 1 \right) \\
  &amp;=   n^{3/2} \left( 1 + \frac{3}{2 n} + \dotsb - 1 \right) \\
  &amp;\ge \frac{3 \sqrt{n}}{2} \\
  &amp; &gt; N
\end{align}$

This is $n &gt; 4 N^2 / 9$. This provides a stride larger than $N$ between lengths of strings in the language, the pumped string will fall short.