2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ what do you mean by solutions? $\endgroup$ – kyary May 4 at 3:48
  • $\begingroup$ Yeah any $x = y > 0$ will do it here. Another example is something like $x = 14, y = 2$ with $14/2 = 7, 15/3 = 5.$ $\endgroup$ – Stephen Donovan May 4 at 3:49
  • $\begingroup$ I apologize I had written the question incorrectly $\endgroup$ – Smartsav10 May 4 at 3:57
  • $\begingroup$ @kyary I have reworded the problem. $\endgroup$ – Smartsav10 May 4 at 4:02
  • $\begingroup$ @ParclyTaxel sorry the question was incorrect before. $\endgroup$ – Smartsav10 May 4 at 4:02
3
$\begingroup$

$16/2$,$15/3$

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

$\endgroup$
2
  • $\begingroup$ May I ask how you came up with that? $\endgroup$ – Smartsav10 May 4 at 4:08
  • $\begingroup$ 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. $\endgroup$ – Eric May 4 at 11:54
3
$\begingroup$

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 $y^2-1$.

$\endgroup$
2
$\begingroup$

\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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.