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

lobht98

lobht98

Answered

2022-06-28

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

massetereqe

massetereqe

Expert

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

Expert

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.

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