# 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.

• what do you mean by solutions? – kyary May 4 at 3:48
• Yeah any $x = y > 0$ will do it here. Another example is something like $x = 14, y = 2$ with $14/2 = 7, 15/3 = 5.$ – Stephen Donovan May 4 at 3:49
• I apologize I had written the question incorrectly – Smartsav10 May 4 at 3:57
• @kyary I have reworded the problem. – Smartsav10 May 4 at 4:02
• @ParclyTaxel sorry the question was incorrect before. – Smartsav10 May 4 at 4:02

$$16/2$$,$$15/3$$

More generally, $$(y^2-1 +1)/y$$ and $$(y^2-1)/(y+1)$$ works.

• May I ask how you came up with that? – Smartsav10 May 4 at 4:08
• I noticed $(-y-1)+1,y$ and $(-y-1),y+1$ worked and adding $y(y+1)$ to the numerator keeps it an integer. Doing it once gives my formula. More generally you can use the Chinese remainder Theorem. – Eric May 4 at 11:54

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 $$y^2-1$$.

\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 $$n+1=12$$.

$$\begin{array}{rrr| rr | rr} m-11 & m & n & x & y & \frac{x+1}y & \frac{x}{y+1}\\ \hline 1 & 12 & 11 & 143 & 12 & 12 & 11 \\ 2 & 13 & 11 & 77 & 6 & 13 & 11 \\ 3 & 14 & 11 & 55 & 4 & 14 & 11 \\ 4 & 15 & 11 & 44 & 3 & 15 & 11 \\ 6 & 17 & 11 & 33 & 2 & 17 & 11 \\ 12 & 23 & 11 & 22 & 1 & 23 & 11 \end{array}$$