Question

Yet more number theory...

"Find all solutions x,y in Z of the Diophantine equation
x^5 + 3*x^2*y^3 + 9*y^5 = 0:

[Hint: Use the method of infinite descent.]

I have no idea how I would even approach this..

Any help would be great! Thanks :)

Answer 1

Hello sarahusher :)

First note that both \(x\) and \(y\) must be even numbers (why?)
Then suppose that \((2x_1,2y_1)\) is a non-zero solution for the equation , putting this in the equation u have\[(2x_1)^5+3(2x_1)^2 (2y_1)^3+9(2y_1)^5=0\]which simplifies to\[x_1^5+3x_1^2y_1^3+9y_1^5=0\]so \((x_1,y_1)\) is a solution to this equation and we must have\[(x_1,y_1)=(2x_2,2y_2)\]using a similar procedure \((x_2,y_2)\) is a solution to the equation and u can create infinite number of solutions\[(x_1,y_1) \\ (x_2,y_2)
\\ . \\ . \\ . \\ (x_n,y_n) \\ . \\ . \\ \]u may notice that \(x_{i+1}<x_i\) and \(y_{i+1}<y_i\) and using method of infinite descent this is a contradiction...so the only solution will be \((x,y)=(0,0)\) ...(why?)

Answer 2

\[\bf 3x^2y^3+9y^5=-x^5 \implies 3y^3(x^2+3y^2)=-x^5 \implies x^2+3y^2=-\frac{ x^5 }{ 3y^3 }\]@mukushla Doesn't re-arranging suggest that \(\bf x,y \ne 0\)?

Answer 3

@mukushla

Answer 4

how u conclude that?

Answer 5

Well the denominator of the right hand side can't be 0 and the denominator. And when x = 0 then the only way the two sides equal each other is when y = 0 hence x and y cannot be 0.

Is my reasoning flawed? Or did I make a mistake in the re-arrangement? @mukushla

Answer 6

@mukushla

Answer 7

im here :) let me think

Answer 8

u say when x=0 the only way is x=y=0 but how about \(x\neq 0\)

Answer 9

@mukushla just put it in wolframalpha and you're that there is only one solution x = 0, y = 0 is correct. But I don't see how that makes sense because when you rearrange, it doesn't work.

Answer 10

I know but saying that x = 0 is the only solution does imply that y must also equal 0. Wolframalpha says thats correct but I don't see how that makes sense once you rearrange.

Answer 11

no u cant say x=0 is the only solution using that re-arrangement

Answer 12

let x>0 and y<0 2 sides are positive..right?

Answer 13

Yes.

Answer 14

Btw I didn't conclude that x = 0, y = 0 is the only solution through re-arrangement. That's what you concluded in your first post. My objection was that once you re-arrange the equation, you see that y cannot be 0. That's what is confusing. However, inserting that in to wolframalpha, what you said initially is correct. Look here:

http://www.wolframalpha.com/input/?i=x%5E5%2B3y%5E3x%5E2%2B9y%5E5+%3D0

@mukushla

Answer 15

oh now i get it...u divide two sides by y^3 thats not allowed because y can take the zero

Answer 16

My bad, looking at wolframalpha more carefully, x = 0, y= 0 is the only integer solution. There is an infinite number of non-integer solutions. However, my only question is why and how is x = 0, y = 0 a solution when in that rearrangement, y clearly can't be 0.

Answer 17

(x,y)=(0,0) is definitely a solution to that equation.

Answer 18

@mukushla Wait, so I made a mistake with the rearrangement? Because we know that the domains for both x and y are all real numbers, I shouldn't be able to divide by 3y^3 since it can also take the value of a 0. Is that your point? @mukushla

Answer 19

yes thats what i meant

Answer 20

you can divide it by 3y^3 provided y not equal to 0.

Answer 21

very right @Abhishek619

Answer 22

mmk

Answer 23

awesome job @mukushla

Answer 24

Thanks oldrin :)

Answer 25

first of all, it is clear that x=0, y=0 is a solution. keeping aside this solution,
now to find the other solutions, divide the whole equation by y^5 provided y not equal to 0.
the equation becomes (x/y)^5+3(x/y)^2+9=0.
let x/y=t .. since the domain of t is the whole real system.
t^5+3t^2=f(t)
f(t)=t^2(t^3+3)
the function has 2 and (-3)^(1/3) as its roots.
now, for f(t)+9, the functional values are just increased by 9 units.
if we plot the graph of t^2(t^3+3) it is as follows.

it has a multiple root at x=o and a root at x=(-3)^(1/3)
also if we add the function by 9 units, we get a graph
there is only one solution at negetive t.
t=-k=x/y
=>x=-ky..hence this proves that there are infinitely many solutions to that equation and adding to that, x=0, y=0 is also a solution.