Given positive integers $a_1\geq a_2\geq a_3$, show that the number of positive integral solutions $(x_1,x_2,x_3)$ of the diophantine equation $$\frac{a_1}{x_1}+\frac{a_2}{x_2}+\frac{a_3}{x_3}=1$$ does not exceed $3a_1a_2(7+2\log a_1)$.

My try:

Clearly we have $x_i>a_i,i=1,2,3$. So let $x_3=a_3+k,x_i=x_i'+a_i,i=1,2$. Then after some algebraic manipulation we have $$(kx_1'-a_1a_3)(kx_2'-a_2a_3)=a_1a_2(a_3+k)^2.$$ Denote by $f(k)$ the number of solutions of this equation and I found this upper bound, using the well-known result that $d(n)=n^{o(1)}$, where $d(n)$ denotes the number of divisors of $n$: $$f(k)\leq \lfloor\frac{d(a_1a_2(a_3+k)^2)}k\rfloor\leq\frac{d(a_1a_2(a_3+k)^2)}k=\frac{(a_1a_2(a_3+k)^2)^{o(1)}}k.$$ Now the number of solutions of the original equation has the upper bound $$\sum_{k=1}^{\infty}\frac{(a_1a_2(a_3+k)^2)^{o(1)}}k,$$ which diverges.

  • $\begingroup$ If $x_i = 3a_i$ isn't it trivial to see that there are infinitely many solutions? $\endgroup$ Commented Aug 30, 2016 at 3:58
  • $\begingroup$ @AlexisOlson $a_i$ are given so they're constant. $\endgroup$
    – Yuxiao Xie
    Commented Aug 30, 2016 at 4:00
  • $\begingroup$ OK. That makes a lot more sense. $\endgroup$ Commented Aug 30, 2016 at 4:02
  • $\begingroup$ The set of solutions really is finite. At least one of the fractions satisfies $a_i/x_i \geq 1/3,$ then you get some factoring business from the other two fractions adding to $1 - (a_i/x_i).$ So you have three cases, the first fraction is at least one third, or the second, or the third. A fair amount of work would be needed to combine that into the desired conclusion. Possibly some inclusion/exclusion, since we could have two of the fractions exceeding 1/3. Suggest you experiment with small $a_1, a_2, a_3, $ get a more specific idea what is going on. I often do computer experiments... $\endgroup$
    – Will Jagy
    Commented Aug 30, 2016 at 18:58

1 Answer 1


Consider the particular solutions of the equation.

For $(a_1,a_2,a_3)=(1,1,1)$ we have ten simple solutions $$(3,3,3), (4,4,2),(4,2,4),(2,4,4),(6,3,2),(6,2,3),(3,6,2),(3,2,6),(2,6,3),(2,3,6).$$ That gives for $(a_1,a_2,a_3)$ in the form of $(x,y,z)$ (i. e. for arbitrary set) ten compound solutions
$$(3x,3y,3z), (4x,4y,2z),(4x,2y,4z),(2x,4y,4z),(6x,3y,2z),(6x,2y,3z),(3x,6y,2z),(3x,2y,6z),(2x,6y,3z),(2x,3y,6z).\qquad (1)$$

For $(a_1,a_2,a_3)=(2,1,1)$ there are ten compound solutions $$(6,3,3), (8,4,2),(8,2,4),(4,4,4),(12,3,2),(12,2,3),(6,6,2),(6,2,6),(4,6,3),(4,3,6).$$ given by $(1)$ for $(x,y,z)=(2,1,1).$

For $(a_1,a_2,a_3)=(2,2,1)$ there are ten compound solutions $$(6,6,3), (8,8,2),(8,4,4),(4,8,4),(12,6,2),(12,4,3),(6,12,2),(6,4,6),(4,12,3),(4,6,6),$$ given by $(1)$ for $(x,y,z)=(2,2,1),$ and simple solution $$(5,5,5).$$ That means that for $(a_1,a_2,a_3)$ in the form of $(2x,2y,z)$ appears additional compound solution $$(5x,5y,5z).\qquad(2)$$

For $(a_1,a_2,a_3)=(2,2,2)$ there are eleven compound solutions $$(6,6,6), (8,8,4),(8,4,8),(4,8,8),(12,6,4),(12,4,6),(6,12,4),(6,4,12),(4,12,6),(4,6,12),(5,5,10),$$ given by $(1)$ for $(x,y,z)=(2,2,2)$ and by $(2)$ for $(x,y,z)=(1,1,2).$

Checks show that for $a\leq2$ given inequality is satisfied.

Then, easy to see that all simple solutions given by condition $$a_1+a_2+a_3\in\mathcal P$$ with $10$ solutions for $(1,1,1)$ and not more that one solution per every another element.

Also note that for every divizor $d\ |\ n,\ d>\sqrt{n}$ exists $d'= n/d,\ d'\ |\ n,\ d'<\sqrt{n},$ so divisors quantity for $n$ can't exceed $2\sqrt{n}+2.$

That means that for any given $a_1$ it exists not more than $2\sqrt{a_1}+2$ divisors of $a_1$, $2\sqrt{a_2}+2$ divisors of $a_2$ and $2\sqrt{a_3}+2$ divisors of $a_3.$

So, if to take in account considered solutions for $a_1\leq2$ and to use the inequality for the averages in the form $$a+b\leq \sqrt{2(a^2+b^2)},$$ then for the element $$(a_1,a_2,a_3),\quad a_1\geq3$$ it can not be more than $$2\sqrt{a_1}(2\sqrt{a_2}+2)(2\sqrt{a_3}+2)+11\leq 8\sqrt{a_1}(\sqrt{a_2}+1)^2+11\leq 8\sqrt{2a_1}(a_2+1)+11 < 12(\sqrt{a_1}a_2+\sqrt{a_1}+1) \leq 12a_1a_2\left(\dfrac2{\sqrt{a_1}}+\dfrac1{a_1}\right)< 9a_1a_2 < 3a_1a_2(7+2\ln{a_1})$$ solutions, and reqired inequality is proved.

  • $\begingroup$ Wouldn't $(2,3,6)$ be a solution too? I don't really understand how your solution works. $\endgroup$
    – Yuxiao Xie
    Commented Sep 5, 2016 at 12:24
  • $\begingroup$ @YuxiaoXie It should be $a_1\geq a_2\geq a_3$ $\endgroup$ Commented Sep 5, 2016 at 16:14
  • $\begingroup$ I mean, for $(a_1,a_2,a_3)=(1,1,1)$, $(2,3,6)$ is also a solution. $\endgroup$
    – Yuxiao Xie
    Commented Sep 6, 2016 at 3:26
  • $\begingroup$ @YuxiaoXie Thanks, fixed. $\endgroup$ Commented Sep 6, 2016 at 6:34

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