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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.

Parcly Taxel
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Smartsav10
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3 Answers3

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$16/2$,$15/3$

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

Eric
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  • May I ask how you came up with that? – Smartsav10 May 04 '21 at 04:08
  • 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 04 '21 at 11:54
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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 $y^2-1$.

Parcly Taxel
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\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}

Steven Alexis Gregory
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