Find all positive integers such that $m+n+mn=2016$

Question

What I have till now is that $2016= 2^5\cdot3^2\cdot 7$.

Also, because $m+n+mn=2016$ then $m$ and $n$ must be even. For the rest my idea is to use congruence module $3$, and $7$ to see all cases. Do you have a better idea? Because there are a lot of cases. How would you find the solutions?

Answer 1

It's $(m+1)(n+1)=2017$ and $2017$ is a prime number.

Thus, $m+1=1$ and $n+1=2017$ or $m+1=2017$ and $n+1=1$, which says that our equation has no solutions.

Answer 2

If you don't spot the factorisation straight away, you should try to separate the variables e.g. try isolating $m$ by writing $ m(n+1)=2016-n$ or $$m=\frac {2016-n}{n+1}$$ Where the right-hand side is an integer. Then divide through so you leave a fraction where the numerator has lower degree than the denominator $$m=\frac {2017}{n+1}-1$$ and you see that if $m$ is an integer, $n+1$ is a factor of $2017$

Answer 3

Hint $\ $ This type of diophantine equation is solvable by a generalization of completing the square. Namely, completing a square generalizes to completing a product, using the AC-method, viz.

$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff\, &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$

So the problem reduces to checking which factors of $\,ad+bc\,$ have above form, a finite process.