I am trying to find a conjecture apparently made by Erdős and Straus. I say...





I am trying to find a conjecture apparently made by Erdős and Straus. I say apparently because I have had so much trouble finding anything information about it that I'm beginning to doubt its existence. Here it is:
Let ϕ ( X ) be a rational function over Q that is defined at every positive integer. If the sum n = 1 ϕ ( n ) converges, it is either rational or transcendental, i.e., it is never an irrational algebraic number.
Has anyone heard of this conjecture? I was told about it by my supervisor, but he doesn't remember where he heard about it.

Answer & Explanation




2022-06-29Added 21 answers

The introduction discusses variations on the following conjecture of Erdős, which apparently goes back to 1949 but did not appear in print in Erdős's intended form until a 1965 paper of Livingston.
Conjecture (Erdős). Let f be a number-theoretic function with period q > 0 such that | f ( n ) | = 1 for 1 n < q and f ( q ) = 0. Then
n = 1 f ( n ) n 0
whenever the series converges.
In 1973, Baker, Birch, and Wirsing attribute the following problem to Chowla:
Problem. Does there exist a rational-valued function f ( n ), periodic with prime period p, such that
n = 1 f ( n ) n = 0 ?
Baker, Birch, and Wirsing disprove the claim by proving the following theorem.
Theorem: Suppose f : Z Q ¯ is a nonvanishing function with period q. If (i) f ( n ) = 0 whenever 1 < gcd ( r , q ) < q, and (ii) the qth cyclotomic polynomial is irreducible over Q ( f ( 1 ) , , f ( q ) ), then
n = 1 f ( n ) n 0.
This result and a related result of Okada are used as a basis for Adhikari, Saradha, Shorey, and Tijdeman to prove the following theorem.
Theorem. Suppose f : Z Q ¯ is periodic with period q. If the series
n = 1 f ( n ) n
converges to some number S, then either S = 0 or S is transcendental.
Applying Baker's theorem tends to lead to dichotomies of the form "S is either rational or transcendental"; beyond that I don't find direct motivation for the claim or prior conjectures of the form you indicate. (But I'm no number theorist.)
Yahir Tucker

Yahir Tucker


2022-06-30Added 8 answers

Assume that Erdos theorem is true. Then f s ( n ) = 1 n 2 s + 1 , s = 1 , 2 , are clearly a rational functions and
n = 1 f s ( n ) = n = 1 1 n 2 s + 1 = ζ ( 2 s + 1 ) ,
will be rationals or transcendentals, which I think is highly considerable statement. Hence I expect the proof to your answer must be quite hard.
Note also that rational function is n 2 2 n 5 + 1 but F n is not, where F n is the n th Fibonacci number.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?