Consecutive Number Divisibility While I was solving a practice problem, I became interested in coming to the conclusion about the following:
Is it possible for both $\frac{x+1}y$ AND $\frac x{y+1}$ to be integers, and if so, how would I find them. Looking at this, I was pretty sure there wasn't any, but I had no concrete mathematical proof. I still don't have a conclusion, which is why I was wondering if any of you all did.
 A: $16/2$,$15/3$
More generally, $(y^2-1 +1)/y$ and $(y^2-1)/(y+1)$ works.
A: For any $y$ there exist infinitely many $x$ with the given equation holding; they are the solutions of $x\equiv-1\bmod y\equiv0\bmod y+1$. Note that $y$ and $y+1$ are coprime, so there is a unique solution modulo $y(y+1)$ by the Chinese remainder theorem, and that is $y^2-1$.
A: \begin{align}
   \dfrac{x+1}y &= m \\
   \dfrac x{y+1} &= n \\
\hline
  x+1 &= my \\
  x &= ny + n \\
\hline
   ny + n + 1 &= my \\
   my - ny &= n+1 \\
\hline
   y &= \dfrac{n+1}{m-n} \\
   x &= n(y+1) \\
\end{align}
So, for example, let $n=11$, then the possible values for $m-11$ are
$1,2,3,4,6,12$, the divisors of $n+1=12$.
\begin{array}{rrr| rr | rr}
   m-11 & m & n & x & y & \frac{x+1}y & \frac{x}{y+1}\\
\hline
   1 & 12 & 11 & 143 & 12 & 12 & 11 \\
   2 & 13 & 11 &  77 &  6 & 13 & 11 \\
   3 & 14 & 11 &  55 &  4 & 14 & 11 \\
   4 & 15 & 11 &  44 &  3 & 15 & 11 \\
   6 & 17 & 11 &  33 &  2 & 17 & 11 \\
  12 & 23 & 11 &  22 &  1 & 23 & 11 
\end{array}
