Prove that there exists no polynomical with integral coefficients such that are satisfied for distinct integers

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- July 22nd 2010, 11:09 AMsashikanthPolynomial in integral coefficients
Prove that there exists no polynomical with integral coefficients such that are satisfied for distinct integers

- July 23rd 2010, 02:06 PMMedia_Mann>2
This problem has captivated me. At first glance, I thought it must be either false or simple. So far, neither seems to be correct. So far all I have managed to prove is the polynomial must be higher than quadratic.

For starters, these "chains" are not difficult to find. Here is a simple example: . . Therefore I figured that the reason no integer "chain" can exist must be numerical, not analytical.

The three points are not collinear (I'll let the reader satisfy him/herself with an explanation for this), therefore cannot be linear. It must have curvature, and therefore must be at least quadratic.

Let and solve using the three sample points in the usual manner.

Using a computer (actually Wolfram Alpha -- shameless plug), one can easily find . The following proves that u is never an integer:

According to Wolfram Alpha,

Substitute , so

So . Call for "difference" and "product" respectively. . Rewriting, . Since the denominator , there is no possible way it can divide the numerator . (Can someone verify this last point rigorously please?)

Since assuming all variables are integers ends in contradiction, one of them must be non-integral. The end result is that must be at least cubic.

I seriously doubt this is the correct approach for this problem, but it is as much as I can contribute, so it may inspire someone else. I believe abstract algebra may hold the key. - July 23rd 2010, 10:07 PMsashikanth
Hi media_man!

Thank you for your post. If it helps, this problem has been picked up from the "Functions" chapter of a 1st year college text book. So I'm reasoning that the solution would be analytic and not numerical. And it would most probably involve some concepts of functions and inverses along with number theory. This book particularly is not very challenging so I think there is a simple concept that I am missing out on. When I first saw it I thought it would be simple too. But I've been grappling with this for over two days to no avail! :( - July 24th 2010, 12:13 AMCaptainBlack
- July 24th 2010, 01:06 AMsashikanth
"Differential Calculus" by Shantinarayan and PK Mittal (S. Chand Publishers). It's an Indian book.

- July 24th 2010, 02:10 AMBourbaki
__Spoiler__: - July 24th 2010, 02:45 AMsashikanth
Thank you so much. This is a beautiful solution!

- July 24th 2010, 02:45 AMtonio
- July 24th 2010, 02:50 AMsashikanth
Use the fact that is divisible by for any positive integer and integers and also use the fact that is a polynomial with integral coefficients to prove that is an integer

- July 24th 2010, 06:54 AMMedia_Man
Excellent proof, Bourbaki. As a continuation, this can also be extended to "chains" of not just three elements, but any arbitrary number. As another continuation, this proves that in any chain (removing the integer requirement), the product of all the elements is always plus or minus 1. Ex: (.347)(-1.879)(1.532)=-1