Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced $m$ in the fractions as $nk$. And after a bit of simple manipulation, I obtained $$k+\frac{1-k}{n+1},$$ $$k+\frac{2-k}{n+2},$$ $$k+\frac{3-k}{n+3},$$ $$k+\frac{4-k}{n+4},$$ $$k+\frac{5-k}{n+5}$$ Now I do not know how to proceed any further.

I have obtained a very interesting form of fraction which will follow all the conditions. Unfortunately I am not seasoned in number theory and Hence my solution is lengthy, confusing and has a large no. Of variable, and has a few assumptions. It would be helpful if someone could suggest a better and shorter solution. The form is (210k+2)/(420k+2) .
