All integers are not expressible as x^n+y^n, help

[Shadow]

During preparations for a mathematical contest I might take part in, I've come across an interesting problem:

 

Let [math] n \in \mathbb{N}[/math] and [math]a, b \in \mathbb{Z}[/math] be integers such that the set [math]\mathbb{Z} \setminus \{ ax^n+by^n, x, y \in \mathbb{Z} \}[/math] is finite. Prove that n = 1.

 

The "official" solution is the ugliest thing I've ever seen, and so incomprehensible that even my professors didn't understand it. My gut was/is telling me that there was a simpler solution, so I've been trying to find one.

I've gotten to the point where all I have to do is prove that the set [math]\mathbb{Z} \setminus \{ x^{2k+1}+y^{2k+1}, x, y \in \mathbb{Z}, k \in \mathbb{N} \}[/math] is finite, ie. that there are infinitely many numbers [math]c \in \mathbb{Z}[/math] that cannot be expressed as [math] x^{2k+1}+y^{2k+1}[/math] for a given k and some integers x, y. And I'm stuck. The only hunch I have is that c will be dependent on k, but other than that, I have nothing. I've been clueless for a week or so now, so I'd really appreciate any ideas any of you might have.

Edited November 23, 2011 by Shadow

[DrRocket]

During preparations for a mathematical contest I might take part in, I've come across an interesting problem:

 

Let [math] n \in \mathbb{N}[/math] and [math]a, b \in \mathbb{Z}[/math] be integers such that the set [math]\mathbb{Z} \setminus \{ ax^n+by^n, x, y \in \mathbb{Z} \}[/math] is finite. Prove that n = 1.

 

The "official" solution is the ugliest thing I've ever seen, and so incomprehensible that even my professors didn't understand it. My gut was/is telling me that there was a simpler solution, so I've been trying to find one.

I've gotten to the point where all I have to do is prove that the set [math]\mathbb{Z} \setminus \{ x^{2k+1}+y^{2k+1}, x, y \in \mathbb{Z}, k \in \mathbb{N} \}[/math] is finite, ie. that there are infinitely many numbers [math]c \in \mathbb{Z}[/math] that cannot be expressed as [math] x^{2k+1}+y^{2k+1}[/math] for a given k and some integers x, y. And I'm stuck. The only hunch I have is that c will be dependent on k, but other than that, I have nothing. I've been clueless for a week or so now, so I'd really appreciate any ideas any of you might have.

 

I have not worked this out in detail, but my initial thought is that one might show that, given a,b and n>1 that for sufficiently large x_0,y_0 and x_1,y_1 the difference between ax_0^n + by^n and ax_1^n+by_1^is greater than 1 (so that there are infinitely many "holes" is the set in question). You can reduce this to the case where a,x_0, x_1,y_0, y_1 are positive and b is negative. Details (and verification that this works) are left to the reader.

[Xittenn]

 

The "official" solution is the ugliest thing I've ever seen, and so incomprehensible that even my professors didn't understand it. My gut was/is telling me that there was a simpler solution, so I've been trying to find one.

 

 

 

Could you post/link this for us?

[Shadow]

Okay, after digging around a little bit I discovered the original solution in its original language, which makes understanding it a lot easier and also cuts away a lot of the ugliness I was perceiving. But it still seems like overkill to me.

 

I have since discovered that the solution I had in mind when I started this topic would be invalid. DrRocket, yours was the approach I was trying to use since I first saw the problem, but a professor of mine shot it down with showing that for a = 1, b = 1, 8 = a*2^2 + b*2^2 and 9 = a*0^2 + b*3^2. Stupidly, it never occurred to me to try and show this applied for almost all integers. Thanks for the idea.