Square-count language is not context-free
= Square-count language is not context-free
The language $\{a^{m^2}b^{m^2}:m\geq0\}$ is not context-free. Pump a word with both block lengths $p^2$. Unequal changes to the blocks destroy equality; an equal positive change $d\leq p$ makes both lengths $p^2+d$, which lies strictly between consecutive squares $p^2$ and $(p+1)^2$.