3
$\begingroup$

A former Olympiad exercise is to find all pairs of integers $(x,y)$ such that $$x(x+1)(x^2+x+2)=2y^2.$$

I could not solve this equation. How can one find all solutions?

I tried to rewrite the above equation in a different form, but it did not help. For example, the equation is equivalent to \begin{align*} x(x+1)(x(x+1)+2)=2y^2&\iff (x(x+1))^2 + 2x(x+1) +1 =2y^2 +1\\ &\iff (x(x+1)+1)^2 =2y^2 +1\end{align*}

Can you tell me how to attack this problem?

Best wishes

$\endgroup$
  • $\begingroup$ I think the best way to start when you have no idea whatsoever is to find some solutions by brute force. They might help you figure out a pattern. $\endgroup$ – Mohammad Zuhair Khan Jan 18 '18 at 5:51
  • $\begingroup$ Do you expect that there are solutions for really large numbers? why/why not? $\endgroup$ – user24142 Jan 18 '18 at 5:52
  • $\begingroup$ Are you allowed to graph for such problems? If so: desmos.com/calculator/vhc57baql1 You can see some answers in the parts close to the origin where some parts of the graph have gradient $1$ or $-1$. $\endgroup$ – Mohammad Zuhair Khan Jan 18 '18 at 5:56
  • $\begingroup$ @Mohammad: Some solutions are easy to see, namely (0,0) and (-1,0). All other solutions must have $y\neq 0$. $\endgroup$ – Oliver Watt Jan 18 '18 at 6:03
  • 1
    $\begingroup$ I have discovered $6$ solutions: $(0,0); (-1,0); (-2, 2); (-2, -2); (1, 2)$ and $(1, -2)$ and I think these are all. $\endgroup$ – Mohammad Zuhair Khan Jan 18 '18 at 6:04
3
$\begingroup$

Set $z=x^2+x$, then we want $z(z+2) = 2y^2$. So $z$ is even and with $w=z/2 \in\Bbb{Z}$, we get the new Diophantine equation $$2w(w+1) =y^2.$$

Notice that since $x^2+x=z$ has no real solution for $z<\frac{1}{4}$, we can assume $w\ge0$.

  • If $2|w$, then since $gcd(2w, w+1)=1$ we have that $w+1$ is a square. Then $2w=z$ is a square, and so is $4z$. But since $x = -\frac{1}{2} \pm\sqrt{z+\frac{1}{4}}$ is an integer, $4z+1$ is a square. So $z=w=0$ which means $y=0$, and the two solutions for $x$ are $0$ and $-1$.

  • If $2$ does not divide $w$, then $gcd(w, 2(w+1))=1$ implies that $w$ is a square, and this implies that $w+1 = 2k^2$ for $k\in \Bbb{Z}$. So $$x^2 +x = 2w = 4k^2-2$$ and since $x = -\frac{1}{2} \pm\sqrt{4k^2-2+\frac{1}{4}} \in \Bbb{Z}$, we have that $16k^2-7$ is a square. So is $16k^2$, and the only integer squares with difference $7$ are $16$ and $9$, so $k=1$, so $w=1$, so $y= \pm2$ and the possibilities for $x$ are $1$ and $-2$.

$\endgroup$
  • $\begingroup$ Your equation $2w(w+1)=y^2$ has infinitely many solutions, for example, $w=49$, $y=70$. The hard part is determining which of these solutions then give solutions for $x^2+x=2w$. $\endgroup$ – David Jan 18 '18 at 6:31
  • $\begingroup$ @David: Right, I was too fast there. I will correct. $\endgroup$ – Torsten Schoeneberg Jan 18 '18 at 6:32
  • $\begingroup$ Good luck!$\,\!$ $\endgroup$ – David Jan 18 '18 at 6:32
  • $\begingroup$ @David: I think I got it now. $\endgroup$ – Torsten Schoeneberg Jan 18 '18 at 9:43
0
$\begingroup$

Since $gcd(x^2+x,x^2+x+2)=2$ we have $x^2+x=2a^2$ and $x^2 +x+2=b^2$ or $x^2+x = a^2$ and $x^2+x+2=2b^2$ where $a,b$ are relatively prime.

1st case: for $x>-2$ we have $$ x^2 <x^2+x+2<(x+1)^2$$ which is impossible. If $x<-1$ then we have $$(x+1)^2 <x^2+x+2<x^2$$ which is again impossible.

2nd case: for $x>0$ we have $$x^2<x^2+x<(x+1)^2$$ which is impossible. So we are left with the case if $x<-1$ we have $$(x+1)^2<x^2+x<x^2$$ impossible. So $x=0$ or $x=-1$...

$\endgroup$
  • $\begingroup$ How do you prove that they're all the solutions? $\endgroup$ – user_194421 Jan 18 '18 at 6:22
  • $\begingroup$ I think your first chain of inequalities holds for $x>1$ (since you need $x+2<2x+1$), hence excludes this case; whereas the second (still in "1st case") holds for $x<-2$ (since you need $x+2<0$), hence excludes that case. Remains indeed the case $x\in\{-2,-1,0,1\}$ which each occur, with the corresponding $y$'s: $\pm2, 0, 0, \pm2$ are immediately calculated $\endgroup$ – Torsten Schoeneberg Jan 19 '18 at 5:34

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.