Proof that a language involving $gcd$ is not context-free
Let $p$ be the pumping length guaranteed by the pumping lemma (for context free languages). Then we choose $m \neq n$ such that $m,n \geq p$ and are both prime. Then clearly $s = a^{n}b^{m} \in L$.

By the pumping lemma we can divide $s$ such that $s=uvxyz$ and


$|vxy| \leq p$
$|vy| \geq 0$
$s' = uv^{i}xy^{i}z \in L$ for all $i \in \mathbb{N}$


For this language we get three similar cases, and one trivial case. The trivial case is where either $v$ or $y$ contains both $a$'s and $b$'s, in which case $s'$ doesn't have the correct ordering, and thus $s'\notin L$.

The nontrivial cases:


 $v$ and $y$ are both strings of $a$'s, then when we pump $s$ we get $s' = a^{n+ik}b^{n}$ where $k\geq 1$. Then $s' \in L$ if $gcd(n+ik,m)=1$, however the modular equation $n+ik\equiv 0 (m)$ has the solution $i \equiv nk^{-1} (m)$, and as $m$ is prime we are guaranteed that $k^{-1}$ exists. Therefore any element of the residue class $i \equiv nk^{-1} (m)$ would give us $gcd(n+ik,m) &gt; 1$. Ergo $s' \notin L$.

Short version: we can pump the string so that $n+ik$ is a multiple of $m$ for any $k$ with $1 \leq k \leq p &lt; m,n$ (which is what we set up at the start).

$v$ and $y$ are both strings of $b$'s, but this case is just the symmetric case to case 1, so we derive a contradiction in this case too.

$v = a^{k}$ and $y = b^{h}$ for some $k$ and $h$ with $1 \leq k+h \leq p$. Then when we pump we get $s' = a^{n+ik}b^{m+ih}$. Now $s' \in L$ if $n+ik \equiv m+ih (m)$ has no solution, but as I'm sure you can see from here, rearranging just gives $n+i(k+h)\equiv 0(m)$, and as before this has solution $i \equiv n(k+h)^{-1}(m)$. So again we derive a contradiction.



Thus there is at least one string in $L$ that cannot be divided as per the pumping lemma and still have all pumping results remain in $L$. Therefore $L$ is not context free. $Q.E.D$

Some additional notes: of course there is an infinite number of strings which can't be pumped, and it's not actually necessary for $m$ and $n$ to be prime, it just simplifies that argument, as we don't have to fiddle around with a prime decomposition - though if you did want to, you just pick one of the prime factors and do the same (or even all of them if you want).