Diophantine Equations

Here is a challenging diophantine problem to ponder about.

2x3=x2y4+9y52x^{3}=x^{2}y^{4}+9y^{5}

6y3=3x3+xy36y^{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
5 years, 11 months ago

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7 votes

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If both xx and yy are rational and nonzero, then r=xyr=\tfrac{x}{y} is rational and nonzero. Since 6y3=3x3+xy36y^3 = 3x^3 + xy^3, we deduce that x=63r3x = 6 - 3r^3, and of course y=xr=63r3ry = \frac{x}{r} = \frac{6-3r^3}{r}. From the first equation we deduce that 2x3=x2y4+9y52x3=x2x4r4+9x5r5  =  x6r4+9x5r52=x3r4+9x2r52r5=rx3+9x2  =  27r(2r3)3+81(2r3)2 \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)  =  27r10162r781r6+2r5+324r4+324r3216r324  =  0 f(r) \; =\; 27r^{10} - 162r^7 - 81r^6 + 2r^5 + 324r^4 + 324r^3 - 216r - 324 \; = \; 0 If we write r=abr = \frac{a}{b} where a,ba,b are coprime integers with b1b \ge 1, then 27a10162a7b381a6b4+2a5b5+324a4b6+324a3b7216ab9324b10  =  0 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 b3b^3 divides 27a1027a^{10}, and hence b3b^3 divides 2727, and hence b=1,3b = 1,3. Moreover aa divides 324b10324b^{10}, and so aa divides 324324.

  1. If b=3b=3 then, since 324=4×81324 = 4\times81 we deduce that aa divides 44. Thus rr must be one of ±13,±23,±43\pm\tfrac13,\pm\tfrac23,\pm\tfrac43. Since none of these six numbers is a zero of ff, this case does not work.

  2. If b=1b=1 then r=ar=a and f(a)=0f(a)=0. Since 2a5=27a(2a3)3+81(2a3)22a^5 = 27a(2-a^3)^3 + 81(2-a^3)^2, we deduce that 33 divides aa. If 323^2 divided aa, then 353^5 would divide f(a)+324=324f(a) + 324 = 324, which it does not. Thus r=ar = a must be one of ±3,±6,±12\pm3, \pm6,\pm12. Since none of these six numbers is a zero of ff, this case does not work either.

There are no nonzero rational roots to these simultaneous equations.

Mark Hennings - 5 years, 11 months ago

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'Thus we deduce that b3 b^3 divides 27a10 27a^{10} ', Can you explain this to me..?? I didnt get this part. Thanks for the solution BTW.. :)

Keshav Gupta - 5 years, 11 months ago

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b3b^3 is a factor of all the other terms.

Mark Hennings - 5 years, 11 months ago

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

Kevin H - 5 years, 11 months ago

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

Mark Hennings - 5 years, 11 months ago

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@Mark Hennings MASTER

Abdullah Farrukh - 5 years, 11 months ago

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@Mark Hennings Thanks Sir! Now i get it

Keshav Gupta - 5 years, 11 months ago

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wow that was fabulos can i meet u in fb

Srikanth Katti - 5 years, 11 months ago

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

Keshav Gupta - 5 years, 11 months ago

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

Kevin H - 5 years, 11 months ago

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

Keshav Gupta - 5 years, 11 months ago

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Lol, x=y=0x=y=0 :P

Michael Tang - 5 years, 11 months ago

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Lol sorry :D x and y≠0

Kevin H - 5 years, 11 months ago

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could you please post the answer to this question,kevin

Vishal Mundra - 5 years, 11 months ago

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there are nonzero, I am waiting for the real problem

said benkadi - 5 years, 11 months ago

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there are nonzero solutions, please give the real problem

said benkadi - 5 years, 11 months ago

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Put x as ky and solve

avinash iyer - 5 years, 11 months ago

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But that would still mean we have 2 variables.

Keshav Gupta - 5 years, 11 months ago

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

Keshav Vats - 5 years, 11 months ago

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@Keshav Vats LOL, hi Keshav.. :P

Keshav Gupta - 5 years, 11 months ago

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@Keshav Gupta :D

Keshav Vats - 5 years, 11 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

said benkadi - 5 years, 11 months ago

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