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Topic: nr + p
Replies: 12   Last Post: Apr 15, 2003 5:52 PM

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Jpr2718

Posts: 397
Registered: 12/8/04
Re: nr + p
Posted: Apr 15, 2003 5:52 PM
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"Eamon " ewarnock@gz.cngb.com wrote:


>Show that, for each r > 1, there are
> infinitely many positive integers
> that cannot be represented in the form
> nr + p.


Perhaps the question is:

Fix r > 1. Show that there are infinitely
many integers s so that s cannot be
written as n^r + p for any integers n and p, n >= 0, and p >= 2 a prime.

I claim that for infinitely many t > 0 the number s = t^r cannot be written
as n^r + p. If n^r + p = t^r, then p = t^r - n^r = (t-n)(t^{r-1} + ... +-
n^{r-1}). Then t-n = 1, so n = t-1, and p = t^r - (t-1)^r. This is a
polynomial in the single variable t, of degree r-1 > 0, and it is known that
any such polynomial takes on composite values infinitely often. When this
polynomial takes a composite value for a given t, t^r cannot be written in the
desired form.

If r is composite, no t^r is n^r + p.

It is not hard to see that when r is prime, t^r - (t-1)^r is irreducible, and
hence is conjectured by Schinzel to take on prime values infinitely often.

It may be that for any sufficiently large s that is not equal to t^q for some q
> 1 that divides r, s can always be written in the desired form.

If r is allowed to vary, then 1 and 11^6 are the only integers up to 10^10 that
are not n^r + p for some n >= 0, r > 1, p >=2 prime (from computer search by
James Van Buskirk). If we require n >= 1, the only additional exceptions are
2, 5, and 1549.

Alessandro Zaccagnini and others have done some work in this area. Some of
Hardy and Littlewood's conjectures are related to the above.

Of course, if the question is something else, the above may be completely
irrelevant.

John Robertson




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