Find all natural numbers $m,n$ such that $m|12n-1$ and $n|12m-1$ Find all natural numbers $m,n$ such that $m|12n-1$ and $n|12m-1$
My progress:
$m,n $ must be coprime.
Also, look  at the expression $12(m+n)-1$ it is divisible by both $m,n$
 A: We can write that, there exist $a,b \in \mathbb{Z}$ such that 
$$12n-1=ma$$ and $$12m-1=nb$$
Or we can say that, $$12mn-m=m^2a$$ and $$12mn-n=n^2b$$
Thus we get that
$$m^2a+m=n^2b+n$$ or $$m(ma+1)=n(nb+1)$$
Since we have already found out that $m,n$ are co-prime, observe that from the above relation,we get 
$$m|(nb+1)$$ and $$n|(ma+1)$$ Or, in other words,
$$nb \equiv -1 \pmod m$$ and $$ma \equiv -1 \pmod n$$
See if this helps in simplifying the problem. 
P.S. This is not a complete solution but it may help you to progress a bit. It was just too long for a comment.
A: Your first observation is very useful, that $\gcd(m,n) = 1$.
The first comment was also very useful, that $m,n \mid 12(m+n)-1$ and hence by the above observation $mn \mid 12(m+n)-1$.
Thus $12(m+n)-1 = kmn$ for some natural $k$, and hence $(km-12)(kn-12) = 144-k$.
Note that $k$ is not divisible by $2$ or $3$.
Since $m,n > 0$ we have $144 > (km-12)(kn-12) \ge (k-12)^2$ and hence $k-12 < 12$.
We now have to check all natural $k$ from $1$ to $23$ that are coprime to $2$ and $3$.
I did a quick check and I think the only solution not mentioned in the first comment comes from $k = 23$ where $(23m-12)(23n-12) = 121$ gives $(m,n) = (1,1)$.
