Pumping lemma for context-free languages

For some $p$, every sufficiently long $w$ in a context-free language has $w=uvxyz$ with $|vxy|\le p$, $|vy|>0$, and $u{v}^{i}x{y}^{i}z$ in the language for all $i\ge 0$.

This formal language is not a context-free language. For pumping length $p$, use $a^p{b}^{p^2}{c}^p$. A pumpable window of length at most $p$ cannot meet both outer blocks. If pumping adds $\alpha ,\beta ,\gamma$ copies of $a,b,c$, then $\alpha \gamma =0$ and $1\le \alpha +\beta +\gamma \le p$. Preserving the product constraint would require $\beta =p(\alpha +\gamma )$, impossible both when $\alpha +\gamma =0$ and when it is positive. This contradicts the pumping lemma for context-free languages.

The language $\{0^p{1}^p:p\text{ is prime}\}$ is not context-free. For a proposed pumping length, start with a larger prime $p$. Pumping unequal numbers of zeros and ones breaks equality; if the positive changes are equal to $d$, pumping with exponent $p+1$ gives the composite common block length $p(1+d)$.

For a word consisting of three long constrained blocks, a pumpable window shorter than each block cannot alter both outer blocks; pumping then breaks a required three-way count relation.

A pumping length is a threshold beyond which every word in the language has a decomposition satisfying the conclusion of the relevant pumping lemma.

The language $\{a^{m^2}{b}^{m^2}:m\ge 0\}$ is not context-free. Pump a word with both block lengths $p^2$. Unequal changes to the blocks destroy equality; an equal positive change $d\le p$ makes both lengths $p^2+d$, which lies strictly between consecutive squares $p^2$ and $(p+1{)}^2$.