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Prove that there are infinitely many integer solutions to the diophantine equation: $(x-y)^7 = x^3y^3$

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    $\begingroup$ "infinite integer" $\: \mapsto \:$ "infinitely many integer" $\;\;\;$ ? $\;\;\;\;\;\;\;$ $\endgroup$ – user57159 Sep 30 '14 at 17:25
  • $\begingroup$ Why is this tagged contest-math? If it's from a contest, please state the contest. :) It's helpful in case some of us want to work more problems like this. $\endgroup$ – apnorton Oct 1 '14 at 4:06
  • $\begingroup$ Will do that next time. Thanks for the reminder $\endgroup$ – Bohan Lu Oct 7 '14 at 2:10
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    $\begingroup$ Why only do it next time? Why not do it now? $\endgroup$ – Joel Reyes Noche May 6 '15 at 2:02
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Let $x=y+h$. Then it's equal:

$$h^7=y^3(y+h)^3$$

Now let's try $y=ah$, you have:

$$h^7=a^3h^3(a+1)^3h^3$$

It's equal:

$$h=a^3(a+1)^3$$

Now you can write down infinitely many integer which satisfies equation:

$$y=a^4(a+1)^3$$

$$x=a^4(a+1)^3+a^3(a+1)^3=a^3(a+1)^4$$

For $a \in \mathbb{N}$.

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  • $\begingroup$ Very intuitive approach. It looks like you may simplify the final answer for 'x'. $\endgroup$ – user121330 Sep 30 '14 at 22:08
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Another solution: Let $x = \frac{k+1}{k} y$, then $$(x-y)^7=x^3 y^3 \Rightarrow \frac{y}{k^7}=\left(\frac{k+1}{k}\right)^3$$

$$\Rightarrow y=k^4(k+1)^3,\qquad x=k^3(k+1)^4$$

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  • $\begingroup$ What motivates making the construction in the first line? $\endgroup$ – Calvin Lin Oct 3 '14 at 23:34
  • $\begingroup$ Intuition and experiment. I noticed that $x=2y$ was a solution but that $x=3y$ or $x=4y$ was not. The next guess I tried was $x=(3/2)y$, then the general pattern. $\endgroup$ – amcalde Oct 4 '14 at 0:16
  • $\begingroup$ That's brilliant! $\endgroup$ – Bohan Lu Oct 7 '14 at 2:28

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