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$k$ is a positive integer. Could we find all the pairs of positive integers $(a,b)$ such that the Diophantine equation $$x(x+a)=y(y+b)$$ has at least $k$ positive integer solutions $(x,y)$, that is $x,y$ are all positive integers?

For example, $a=2(k+1)+\dfrac{(2k+2)!!}{2k+2},b=\dfrac{(2k+2)!!}{2k+2}-2(k+1)$. I find it luckily , without System approach

This problem's background is the 2014 China Mathematical Olympaid P2 problem. The exam is only want a pair $(a,b)$. So, I want to find all pairs $(a,b)$.

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  • $\begingroup$ Do you know how to find the number of solutions for some given $a, b$? $\endgroup$ – Karolis Juodelė Dec 26 '13 at 22:03
  • $\begingroup$ Since there are infinitely many such pairs, what do you mean by "find"? $\endgroup$ – Robert Israel Dec 26 '13 at 23:12
  • $\begingroup$ @RobertIsrael analytic expression give all such $a,b$ ? $\endgroup$ – ziang chen Dec 27 '13 at 0:03
  • $\begingroup$ Could you give us some background or context for this question, such as what you have tried, where you have got stuck, and where this question comes from? $\endgroup$ – Old John Dec 30 '13 at 23:25
  • $\begingroup$ @OldJohn Thanks a lot! I have edit the problem $\endgroup$ – ziang chen Dec 30 '13 at 23:29
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$x(x+a) = (x+\frac a2)^2-\frac{a^2}4$, so $x(x+a)=y(y+b)$ if and only if $$\left(x+\frac a2\right)^2-\left(y+\frac b2\right)^2=\frac{a^2-b^2}{4}$$ if and only if $$(2x+2y+a+b)(2x-2y+a-b)=(a+b)(a-b). $$ So if we let $c=a+b, d=a-b$, then any factorization of $cd$ into factors $c'd'$ of the same parity (i.e. $c'\equiv c\pmod 2$, $d'\equiv d\pmod 2$) gives rise to the solution $x=\frac{c'-c+d'-d}{2}$, $y=\frac{c'-c+d-d'}{2}$. And if $c'>c$ and $c'-c>|d'-d|$ these solutions are positive, and if $c>|d|\ge0$ then also $a,b>0$.

For $k>0$ let $c=3^{k+1}$, $d=3^{k}$. Then for $0< i\le k$ we can let $c'=3^{k+1+i}$, $d'=3^{k-i}$, that is with $$ a=\frac{3^{k+1}+3^{k}}2=2\cdot 3^{k}, \qquad b=\frac{3^{k+1}-3^{k}}2=3^{k}$$ we have at least the $k$ solutions $$ x=\frac{3^{k+1+i}-3^{k+1}-3^{k-i}+3^k}{2}, \qquad y=\frac{3^{k+1+i}-3^{k+1}+3^{k-i}-3^k}{2}$$

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  • $\begingroup$ Of course $3$ can be replaced by any odd integer $\ge 3$. Another special case is $a=b$, where you have infinitely many positive integer solutions $x=y$. $\endgroup$ – Robert Israel Dec 26 '13 at 23:15

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