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All integers are not expressible as x^n+y^n, help Rate Topic: -----

#1 23 November 2011 - 01:56 PM Shadow 


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

Let n \in \mathbb{N} and a, b \in \mathbb{Z} be integers such that the set \mathbb{Z} \setminus \{ ax^n+by^n, x, y \in \mathbb{Z} \} 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 \mathbb{Z} \setminus \{ x^{2k+1}+y^{2k+1}, x, y \in \mathbb{Z}, k \in \mathbb{N} \} is finite, ie. that there are infinitely many numbers c \in \mathbb{Z} that cannot be expressed as x^{2k+1}+y^{2k+1} 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.

This post has been edited by Shadow: 23 November 2011 - 01:58 PM

This applies to the above post and any new ideas or media presented in it.

- "Cryptographically secure linear feedback shift register based stream ciphers" -- a phrase that'll get any party started.
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#2 23 November 2011 - 04:09 PM DrRocket 


Primate

View PostShadow, on 23 November 2011 - 01:56 PM, said:

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

Let n \in \mathbb{N} and a, b \in \mathbb{Z} be integers such that the set \mathbb{Z} \setminus \{ ax^n+by^n, x, y \in \mathbb{Z} \} 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 \mathbb{Z} \setminus \{ x^{2k+1}+y^{2k+1}, x, y \in \mathbb{Z}, k \in \mathbb{N} \} is finite, ie. that there are infinitely many numbers c \in \mathbb{Z} that cannot be expressed as x^{2k+1}+y^{2k+1} 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.

You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... -- Richard P. Feynman
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#3 23 November 2011 - 05:49 PM Xittenn 


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View PostShadow, on 23 November 2011 - 01:56 PM, said:


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?
"He is their god! He leads them like a thing made by some other deity than Nature that shapes man better. And they follow him against us brats with no less confidence than boys pursuing summer butterflies, or butchers killing flies." - Cominius; Shakespears Coriolanus
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#4 3 December 2011 - 12:10 PM Shadow 


Atom
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.
This applies to the above post and any new ideas or media presented in it.

- "Cryptographically secure linear feedback shift register based stream ciphers" -- a phrase that'll get any party started.
0

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