0
$\begingroup$

Find every pair of integers $(p,q)$ such that $ \dfrac {p^3 -q}{pq+1} $ is an integer.

I have tried few pairs, and came with few answers. But I couldn't come with proper solution.

$\endgroup$
  • $\begingroup$ What are the solutions that you found? $\endgroup$ – TZakrevskiy Aug 22 '16 at 8:32
  • 2
    $\begingroup$ The two statements (title & body) you've given are different. Which one is correct? $\endgroup$ – Matt Aug 22 '16 at 8:32
  • $\begingroup$ I corrected it. Thanks. $\endgroup$ – Tahir Imanov Aug 22 '16 at 8:33
  • $\begingroup$ @TZakrevskiy $(q^3 ,q) $ as an example. $\endgroup$ – Tahir Imanov Aug 22 '16 at 8:35
  • 5
    $\begingroup$ $$\dfrac{q(p^3-q)}{pq+1}=p^2-\dfrac{p^2+q^2}{pq+1}$$ See math.stackexchange.com/questions/28438/… $\endgroup$ – lab bhattacharjee Aug 22 '16 at 8:37
1
$\begingroup$

Take any integer $n$. Consider a linear recurrent sequence: $$a_0=0$$ $$a_1=n$$ $$a_{i+1}=n^2\cdot a_i-a_{i-1}$$ The first few sequences are: $$\{0,2,8,30,112,418,1560,5822,21728\dots\}$$ $$\{0,3,27,240,2133,18957,168480,1497363\dots\}$$ $$\{0,4,64,1020,16256,259076,4128960\dots\}$$ Then any pair of subsequent terms (in any order) is a solution. Note also that if we take $(a_i,a_{i+1})$ as our $(p,q)$, the integer they produce is $a_{i-1}$, and if we consider $(p,q)=(a_{i+1},a_i)$ instead, we get $a_{i+2}$.

I'm not sure whether this covers all solutions, though. But I strongly suspect it does.

$\endgroup$
  • 1
    $\begingroup$ You could start with $a_0=0$ to make it simpler and to include the trivial solution p=0. $\endgroup$ – Jaap Scherphuis Aug 22 '16 at 9:35
  • $\begingroup$ Yeah, you are right. $\endgroup$ – Ivan Neretin Aug 22 '16 at 9:36
1
$\begingroup$

Let $p,q,r\in \mathbb{Z}$

$$\frac{p^3-q}{pq+1}=r \iff \frac{p^3-r}{pr+1}=q$$

Note that $q$ and $r$ are symmetric in roles,

take $q=a_{k+1}$, $r=a_{k-1}$ with $p=a_{k} \,$,

$$a_{k+1}=\frac{a_{k}^3-a_{k-1}}{a_{k} a_{k-1}+1} \iff a_{k-1}=\frac{a_{k}^3-a_{k+1}}{a_{k} a_{k+1}+1}$$

Take $a_{0}=0$ and $a_{1}=n$, that reproduces what Ivan Neretin found.

In particular, for $n=1$, $$a_{k}=\frac{2}{\sqrt{3}} \sin \frac{\pi k}{3}$$

And for $n>1$, $$a_{k}=\frac{n}{\sqrt{n^4-4}} \left[ \left( \frac{n^2+\sqrt{n^4-4}}{2} \right)^k- \left( \frac{n^2-\sqrt{n^4-4}}{2} \right)^k \right]$$

$\endgroup$

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.