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# 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
3 years, 10 months ago

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there are nonzero solutions, please give the real problem · 3 years, 10 months ago

there are nonzero, I am waiting for the real problem · 3 years, 10 months ago

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. · 3 years, 10 months ago

wow that was fabulos can i meet u in fb · 3 years, 10 months ago

'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.. :) · 3 years, 10 months ago

$$b^3$$ is a factor of all the other terms. · 3 years, 10 months ago

Thanks Sir! Now i get it · 3 years, 10 months ago

This is wonderful! How did you come up with this? Are you some sort of math professor? · 3 years, 10 months ago

I've been doing Maths for a while. This problem is fiddly, but did not need tough ideas to solve. · 3 years, 10 months ago

MASTER · 3 years, 10 months ago

could you please post the answer to this question,kevin · 3 years, 10 months ago

'Find x and y so that they are rational'.. but isnt Diophantine Equation about integer solutions? · 3 years, 10 months ago

Thank you for bringing this up I may need to clarify this problem. · 3 years, 10 months ago

You're Welcome.. but this is a good question anyways.. I'm still working on it! :P · 3 years, 10 months ago

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 · 3 years, 10 months ago

Put x as ky and solve · 3 years, 10 months ago

But that would still mean we have 2 variables. · 3 years, 10 months ago

Also, k will be a rational number to maintain generality, possibly complicating the question. · 3 years, 10 months ago

LOL, hi Keshav.. :P · 3 years, 10 months ago

:D · 3 years, 10 months ago

Lol, $$x=y=0$$ :P · 3 years, 10 months ago