None of this nursery school stuff - a proper maths problem. 25$ reward

[Rotating Rabbit]

X^2 + 68 = Y^5

Find all the solutions or prove there are none.

25$ via neteller for the first correct answer.

(So if you have finite solutions you must list and show they solve the equation and also prove no others work. Or if you have an infinite set X of solutions you must show that all solutions from set X are roots, and prove no others work. Or you must prove there are no solutions)

[ihardlyknowher]

There are infinite solutions, since every number > or = 68 has a defined 5th root.

In other words, let X = x and Y=(x^2+68)^(1/5).

The has a solution for any integer x.

[jason_t]

[ QUOTE ]
X^2 + 68 = Y^5

Find all the solutions or prove there are none.

25$ via neteller for the first correct answer.

(So if you have finite solutions you must list and show they solve the equation and also prove no others work. Or if you have an infinite set X of solutions you must show that all solutions from set X are roots, and prove no others work. Or you must prove there are no solutions)

[/ QUOTE ]

I think that you intended to state the solutions should be natural numbers otherwise the problem is uninteresting.

[jason_t]

[ QUOTE ]
There are infinite solutions, since every number > or = 68 has a defined 5th root.

In other words, let X = x and Y=(x^2+68)^(1/5).

The has a solution for any integer x.

[/ QUOTE ]

I think he intended for both x and y to be natural numbers.

[gumpzilla]

It's supposed to be a Diophantine equation, he wants solutions for integer X and Y. Doesn't seem like the most thrilling problem in the world, and involves more technical knowhow than counting triangles.

[Rotating Rabbit]

Yeah sorry I (obviously) meant X and Y natural numbers. And I intend to pay up unlike KKF btw.

[MrWookie47]

Y = (X^2+68)^(1/5) describes a curve with a domain of the entire set of real numbers. There are infinite solutions. However, since this is a 5th root, there are 4 additional solutions that each yield the same value of y when one considers numbers in the imaginary plane. The set of roots in the imaginary plane, when described in polar coordinates, will have the same radius as the real solution, and will have their angles spaced equidistantally around 2*pi radians. The set of all roots can be described by

(x^2 + 68)^(1/5) * exp(i * 2*pi*n/5)

where n is 1, 2, 3, 4, or 5 (or any integer, I suppose), i is the sqrt(-1), and x is any real number.

Edit: Crap. I guess this problem is hard because he wanted the natural number solutions, not the imaginary ones.

[jason_t]

[ QUOTE ]

Edit: Crap. I guess this problem is hard because he wanted the natural number solutions, not the imaginary ones.

[/ QUOTE ]

Yes. Diophantine equations, of which this is an example, are extremely difficult to solve. Fermat's last theorem is another example. A famous problem of Hilbert, that has been solved, qualifies exactly how difficult they are to solve: there will never ever be a computer algorithm capable of solving all Diophantine equations.

[TheIrishThug]

x^2+68-y^5
x=+/- (y^5-68)^1/2
y has to be >= 68^1/5
x can be any real number

edit: so there r an infinite solution, but with the range of y>=68^1/5

[Patrick del Poker Grande]

[ QUOTE ]
the problem is uninteresting.

[/ QUOTE ]