From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.num-analysis,sci.math
Subject: Re: Integer roots
Date: 28 Feb 1998 09:52:27 GMT
Juan Martin wrote:
>I want know if in the equation :
>
>(2i+a)^n-2(i+a)^n-a^n=0
>
>a and 2i can be both integers positives.
Perhaps no one has answered this here because this is a question in number
theory, not numerical analysis. Newsgroups like sci.math would be better.
(Follow-ups set.)
I suspect the answer to your question is that there are no such positive
integers a,i for any integer n. I can recast the question, into a form
which probably admits an answer; but I didn't complete the investigation.
Setting x = i/a in your equation shows you are simply looking for a positive
rational solution x to the equation
(2x+1)^n - 2 (x+1)^n - 1 = 0
Note that all coefficients are even integers, so we divide by two. Then
by the rational root test, any such x has numerator +-1 and denominator
dividing N = 2^(n-1) - 1. This is easy enough to test for any n: try all
divisors d of N and see if any has a reciprocal which is a root x.
We can be more efficient: let
f(x) = (2x+1)^n - 2 (x+1)^n - 1
so we seek positive zeros of f. Now, f'(x) = 2n(2x+1)^(n-1) - 2n(x+1)^(n-1)
is never zero except when x=0, and is positive if x=1, so f is increasing on
(0, oo). Clearly f is negative when x=0 and approaches +oo when x does, so
f has precisely one positive real root r.
We can estimate r: letting x = y/n we see we are asking for solutions to
(2y/n+1)^n - 2 (y/n+1)^n - 1 = 0;
for large n these terms are roughly exp(2y) and exp(y), respectively,
so we have to solve a quadratic in exp(y), and so see y must be around
ln(1+sqrt(2)); that is, the root r is approximately
ln(1+sqrt(2)) / n
That isn't really a fair analysis, but the conclusion is valid;
indeed, an asymptotic formula for the root r can be obtained:
r = ln(1+sqrt(2)) / n + (4+sqrt(2))ln(1+sqrt(2))^2 / 4n^2 + O(1/n^3)
Thus the number d = 1/r which must be an integral divisor of N=2^(n-1) - 1
is approximately n/ln(1+sqrt(2)); more precisely, it's
d = a n + b + c/n + O(1/n^2)
where
a = 1/ln(1+sqrt(2)) = 1.13459
b = -(1 + sqrt(2)/4) = -1.35355
c = (1/a)*(20 - 3sqrt(2)/a)/96 = 0.14929
Of course, the original question only has an affirmative answer if this
number d is an _integer_, which must then be the nearest integer to
R = a n + b
(for large n, e.g. n> 2 !) and in fact must be a little greater than R.
So it's now trivial to check for the rationality of roots of f:
1. Compute R above
2. If ceiling(R)-R isn't just over 0, then f has no rational roots.
3. Otherwise, compute d=round(R); this divides N iff f has rational roots.
I carried out the computations for n up to about 100000. Interestingly,
I never got to step 3; that is, R was never close enough to an integer.
We are only interested in cases in which d is an integer, so that R
is just c/n + O(1/n^2) shy of an integer, i.e., the quantity
S=c/(ceiling(R)-R)/n will have to be about 1 (really 1 + O(1/n); with
some effort I'm sure this could be transformed into a statement like
"if d is an integer than S > 0.9, for n >> 0". )
Well, in some cases, R came fairly close to being an integer, but S was
never very close to 1. I found examples (n,S)=(62, 0.27), (545, 0.49), and
(28444, 0.36), but a search up to n=100000 found no other S>0.2, and only a
handful of cases which could even muster S>0.1 (namely n=114, 597, 1392,
2239, 29291, 80059).
To prove that we never have S > 1/2, it would suffice to show that
there are no pairs of integers (n,m) for which | a n + b + m | < 2c/n.
(A few known small cases can be excluded.) I don't remember enough of
Transcendental Number Theory to know whether this is true, although I
suppose a careful reading of Baker's book would help.
Of course it's quite possible I've overlooked some simple divisibility
condition, for example, which would preclude having f(1/d)=0 for some
integral divisor d of N=2^(n-1)-1.
dave