From: email@example.com (Stefan Bartels) Newsgroups: sci.math.research Subject: Number theory proof of Curtiss is wrong Date: 16 Jul 1996 10:52:23 GMT Organization: Rechenzentrum Universitaet Kiel
Did anybody ever realize a bad mistake in the proof of Curtiss in his article "On Kellogg's Diophantine Problem", Amer. Math. Monthly, 29 (1922), pp. 380-387?
I don't think so, because Richard K. Guy in his book "Unsolved Problems in Number Theory", Springer 1994, pp. 162f., and even Erdös in his article "Comment on problem E2427", Amer. Math. Monthly, 81 (1974), pp. 780-782, not only cite Curtiss' article in their bibliography but also mention in their text explicitly him to have proved the result.
By Kellogg was the question raised, what the greatest denominater in a representation of 1 as a sum of k unit-fractions of integers (1 = sumkn=1 1/xn) is. He conjectured for the maximum of the xn: x1 = 1, xk+1 = xk (xk + 1). This result was "proved" by Curtiss.
The mistake mentioned below is just concerning the proof and not the theorems of Curtiss, because Takenouchi in his article "On an Indeterminate Equation", Proc. of the Physico-Mathematical Society of Japan, 3rd series, vol. 3, pp. 78-92, proved something similar. Also Erdös in his article mentioned above says that he has proved a generalization of the theorem in his Hungrian article "On a Diophantine Equation", Mat. Lapok, 1 (1950), pp. 192-210.
To prove the theorem by the methods introduced by Curtiss one need (as I think and as I did) a lot more space than he did.
To understand the following you have to have the original article. I use "g" instead of "\varphi".
The mistake is on page 385. On page 384 he states the Theorem III. To prove the theorem he starts with any set satisfying equation (11) and "shows" that a transformation by his second method leads to a reduced set which also satisfies equation (11). This statement is wrong: Consider the set (3,3,4), which is obviously reduced. Thus the set (3,3) is satisfying equation (11). By his second method he gains the set (2,6) which does not satisfy equation (11) for fn-2(x) = 3 < 6!! (He should have taken instead the set (2,4) for gn-2(3,3) = 36 < 40 = gn-2(2,4)! Even if you consider the whole new set (2,4,6), it is impossible to prove what he "did" that Xn-2/2 ( Xn-2/2 + 1) <= fn-2(x), for 4,6 > 3 = fn-2(x)! Thus the real problem on how to gain a new set satisfying equation (11) and increasing gn-2 from any set satisfying equation (11) is not solved by Curtiss.)
His mistake is in the line "But since X is compact ...". He has two sets X. The original set (x1, ... xn-1) was compact, but the shortened set (x1, ... xn-2) is **not** always compact! This is the case with the set (3,3,4) from above.
Thus the Theorem III is not proven by Curtiss.
From: firstname.lastname@example.org (Bartels) Date: 24 Oct 1996 15:06:25 GMT Newsgroups: sci.math.research Subject: Maximums of sums of Egyptian fractions
In the book
Paul Erdös & Ronald L. Graham: "Old and New Problems and Results in Combinatorial Number Theory", L'Engseignement Mathematique Universite de Geneve, 1980
one can read on page 31:
It is true that for any rational a/b, the closest strict underapproximation Rn(a/b) of a/b by a sum of n unit fractions is given by
Rn(a/b) = Rn-1(a/b) + 1/m
where m is the least denominator not yet used for which Rn(a/b) < a/b, provided that n is sufficiently large.
The authors do not give any citation of the proof of this theorem. Does anybody know in which article it is proven or maybe know how it can be proven?
From: email@example.com (Gerry Myerson) Date: 25 Oct 1996 06:19:23 GMT Newsgroups: sci.math.research Subject: Re: Maximums of sums of Egyptian fractions
In the case where the rational you are trying to approximate is 1 (one), see D. R. Curtiss, On Kellogg's diophantine problem, Amer. Math. Monthly 29 (1922) 380--387. However, Curtiss' proof, while entirely elementary, is not easy to follow and, according to a post to this group earlier this year, is incorrect.
A simpler proof is given by O. Izhboldin and L. Kurliandchik, Unit fractions, Proc. St. Petersburg Math. Soc. 3 (1995) 193--200. This appears in English translation in Amer. Math. Soc. Translations, Series 2, 166.
They say nothing about the general rational, and I have no opinion as to whether their methods generalize.
Gerry Myerson (firstname.lastname@example.org)
From: email@example.com (Bartels) Date: 30 Oct 1996 11:23:49 GMT Newsgroups: sci.math.research Subject: Re: Maximums of sums of Egyptian fractions
This posting is a completion to the posting of Gerry Myerson (firstname.lastname@example.org) with two purposes:
Gerry mentioned a "former posting" saying that the proof of Curtiss is wrong. This posting was submitted to this newsgroup by me in July. In the meantime I found out that the proof is definitely right (but obviously difficult to understand). (Sorry for that.) But this mistake (of me) led to a generalization of Curtiss' theorem that has not been published elsewhere yet: The best underapproximation of a rational a/b is given by the greedy algorithm at least for the case b = -1 mod a. (In the case a=b=1 this is Curtiss' theorem; in the case a=1 this is a statement proved by Erdös in 1950.)
I was also able to prove that the methods Curtiss invented work if and only if b = -1 mod a. But there seem to be a lot more rationals for which the underapproximation is given by the greedy algorithm. While trying to characterize these rationals I came across the theorem of Erdös and Graham, which I mentioned in my first posting on this subject.
Any hint on the proof of this theorem would be helpful.