There is a problem which I cannot solve. If you give a tip I will be very glad.Prove that following language is not context free:$L= \{ a^nb^m | \gcd(n,m) = 1 \}$.It can be proven using the pumping lemma, but how?If I start with some prime numbers $m$ and $n$ where $m>n>2$ and pump it up from $uVxYz$, there are three possible outcomes: $a^{n + k} b^m$, $a^{n +k}b^{m +k}$, $a^n b^{m +k}$. Since I do not know whether $k$ is even or odd I cannot say something. It is certain that $a^n$ and $b^m$ will be odd. However after adding $k$ to some of them, how can I say something about whether their gcd is 1 or not?
A context free grammar proof
formal languages;context free;pumping lemma