Prove that there exist infinitely many pairs $(x,y)$, where $x$ and $y$ are positive integers, such that $(x-y)^7=x^3 y^3$

I've already got the solution in comment, but is there detailed procedure and detailed method to get that solution?

  • $\begingroup$ I don't think so. Wolfram gives only $2$. $\endgroup$ – Aditya Agarwal Oct 15 '15 at 14:03
  • $\begingroup$ Try $(x,y)=(a^3(a+1)^4, a^4(a+1)^3), \ a\in \Bbb{Z}$ $\endgroup$ – wwwrqnojcm Oct 15 '15 at 14:09
  • $\begingroup$ @wwwrqnojcm: Yes, that looks like a solution. What is your question, then? $\endgroup$ – Henning Makholm Oct 15 '15 at 14:13
  • $\begingroup$ So, you've found infinitely many solutions, @wwwrqnojcm? Aren't you done? $\endgroup$ – Thomas Andrews Oct 15 '15 at 14:32
  • $\begingroup$ But is there detailed procedure to get that solution? I mean what is the method to find that pair of (x,y)? I also posted this question on AoPS but only got an answer instead of detailed procedures $\endgroup$ – wwwrqnojcm Oct 15 '15 at 14:35

Well, here is what I did to get the same answer:

Define the positive integer $d=x-y$. We can rewrite the equation $(x-y)^7=x^3y^3$ as: $$ d^7 = (d+y)^3y^3 $$ From this we can see that $y$ must be a multiple of $d$ (since the right-hand-side must be a multiple of $d$). So we can write $y=ad$ for some positive integer $a$. From that we get: $$ d^7 = (d+ad)^3a^3d^3 $$ Simplifying gives $d=a^3(1+a)^3$. So we can choose $a$ to be any positive integer we like, and then define $d=a^3(1+a)^3$. The result is the same as your answer. This also proves that it is the only possible answer.


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