Prove that the following language is not a regular language
Assume towards contradiction that the language is regular.
Let a be the &quot;pumping length&quot; given by the pumping lemma.
Let $p \geq a+2$ be a prime number. Consider the word: $w=0^p1^p \in L$.
By the pumping lemma, it can be written as $w=xyz$ where $|y| \geq 1$,
$|xy| \leq a $, $xy^nz \in L \ \ ,\forall n \geq 0$.
Consider $w' = xy^0z$.
Note that the length of zeros in w' is strictly less than p, and that the length of ones in w' is still exactly p. Also note that the length of zeros in w' is at least 2 since $z$ itself contains at least two copies of zero. Therefore, the number of zeros in w' is $ 1 &lt; k &lt; p $.
But in order to be in $L$, $w'$ has to satisfy $p \ \ mod \ \ k = 0$, which is equivalent to $k | p$ - contradiction!