For how many pairs of positive integers n and m is the statement $mn - 8m + 6n =0$ true? 
Each interior angle of a regular polygon with $n$ sides is $\frac{3}{4}$
  of each interior angle of a second regular polygon with $m$ sides.
How many pairs of positive integers $n$ and $m$ are there for which this
  statement is true?

$\frac{(n-2)*180}{n}$ is the value of one interior angle for a polygon with $n$ sides.
Therefore $\frac{(n-2)*180}{n} =\frac{3(m-2)*180}{4m}$ and $mn - 8m + 6n =0$
For how many pairs of positive integers $n$ and $m$ is the statement $mn - 8m + 6n =0$ true?

$$mn-8m+6n=0\implies n(m+6)=8m\implies n=\frac{8m}{m+6}$$
Testing this formula from $1\le m\le 1000$, the only integer solutions found were
$$\frac{2\times8}{2+6}=2
\quad\frac{6\times8}{6+6}=4
\quad\frac{10\times8}{10+6}=5
\quad\frac{18\times8}{18+6}=6
\quad\frac{42\times8}{42+6}=7$$
If we solve the other way around,  $$m=\frac{6n}{8-n}$$ and we can see a limit where $1\le n\le7$ so the pairs are $(2,2), (6,4), (10,5), (18,6), (42,7)$.
