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

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## Comments

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TopNewestIf 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 \(324\).

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.

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.

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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.. :)

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\(b^3\) is a factor of all the other terms.

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wow that was fabulos can i meet u in fb

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

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

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

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there are nonzero solutions, please give the real problem

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there are nonzero, I am waiting for the real problem

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could you please post the answer to this question,kevin

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Lol, \(x=y=0\) :P

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Lol sorry :D x and y≠0

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Put x as ky and solve

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But that would still mean we have 2 variables.

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

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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

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