# Diophantine Equations

Here is a challenging diophantine problem to ponder about.

$2x^{3}=x^{2}y^{4}+9y^{5}$

$6y^{3}=3x^{3}+xy^{3}$

Find x and y so that they are integer solutions*

• x and y are not equal to 0 Note by Kevin H
7 years, 7 months ago

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## Comments

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If both $x$ and $y$ are rational and nonzero, then $r=\tfrac{x}{y}$ is rational and nonzero. Since $6y^3 = 3x^3 + xy^3$, we deduce that $x = 6 - 3r^3$, and of course $y = \frac{x}{r} = \frac{6-3r^3}{r}$. From the first equation we deduce that $\begin{array}{rcl} 2x^3 & = & x^2y^4 + 9y^5 \\ 2x^3 & = & x^2 \frac{x^4}{r^4} + 9\frac{x^5}{r^5} \; = \; \frac{x^6}{r^4} + \frac{9x^5}{r^5} \\ 2 & = & \frac{x^3}{r^4} + \frac{9x^2}{r^5} \\ 2r^5 & = & rx^3 + 9x^2 \; = \; 27r(2 - r^3)^3 + 81(2-r^3)^2 \end{array}$ and hence $f(r) \; =\; 27r^{10} - 162r^7 - 81r^6 + 2r^5 + 324r^4 + 324r^3 - 216r - 324 \; = \; 0$ If we write $r = \frac{a}{b}$ where $a,b$ are coprime integers with $b \ge 1$, then $27a^{10} - 162a^7b^3 - 81a^6b^4 + 2a^5b^5 + 324a^4b^6 + 324a^3b^7 - 216ab^9 - 324b^{10} \; = \; 0$ Thus we deduce that $b^3$ divides $27a^{10}$, and hence $b^3$ divides $27$, and hence $b = 1,3$. Moreover $a$ divides $324b^{10}$, and so $a$ divides $324$.

1. If $b=3$ then, since $324 = 4\times81$ we deduce that $a$ divides $4$. Thus $r$ must be one of $\pm\tfrac13,\pm\tfrac23,\pm\tfrac43$. Since none of these six numbers is a zero of $f$, this case does not work.

2. If $b=1$ then $r=a$ and $f(a)=0$. Since $2a^5 = 27a(2-a^3)^3 + 81(2-a^3)^2$, we deduce that $3$ divides $a$. If $3^2$ divided $a$, then $3^5$ would divide $f(a) + 324 = 324$, which it does not. Thus $r = a$ must be one of $\pm3, \pm6,\pm12$. Since none of these six numbers is a zero of $f$, this case does not work either.

There are no nonzero rational roots to these simultaneous equations.

- 7 years, 7 months ago

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'Thus we deduce that $b^3$ divides $27a^{10}$ ', Can you explain this to me..?? I didnt get this part. Thanks for the solution BTW.. :)

- 7 years, 7 months ago

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$b^3$ is a factor of all the other terms.

- 7 years, 7 months ago

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This is wonderful! How did you come up with this? Are you some sort of math professor?

- 7 years, 7 months ago

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I've been doing Maths for a while. This problem is fiddly, but did not need tough ideas to solve.

- 7 years, 7 months ago

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MASTER

- 7 years, 7 months ago

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Thanks Sir! Now i get it

- 7 years, 7 months ago

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wow that was fabulos can i meet u in fb

- 7 years, 7 months ago

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'Find x and y so that they are rational'.. but isnt Diophantine Equation about integer solutions?

- 7 years, 7 months ago

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Thank you for bringing this up I may need to clarify this problem.

- 7 years, 7 months ago

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You're Welcome.. but this is a good question anyways.. I'm still working on it! :P

- 7 years, 7 months ago

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Lol, $x=y=0$ :P

- 7 years, 7 months ago

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Lol sorry :D x and y≠0

- 7 years, 7 months ago

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could you please post the answer to this question,kevin

- 7 years, 7 months ago

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there are nonzero, I am waiting for the real problem

- 7 years, 7 months ago

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there are nonzero solutions, please give the real problem

- 7 years, 7 months ago

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Put x as ky and solve

- 7 years, 7 months ago

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But that would still mean we have 2 variables.

- 7 years, 7 months ago

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Also, k will be a rational number to maintain generality, possibly complicating the question.

- 7 years, 7 months ago

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LOL, hi Keshav.. :P

- 7 years, 7 months ago

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:D

- 7 years, 7 months ago

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what do ypu want exactely, to solve problem ou to do something, like jocking how many solotion do you want wat is the problem please, i need more details for discussing with you i am from morroco nice to know and meet you

- 7 years, 7 months ago

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