Putnam 2017, Problem B-3

Suppose that \( f(x) = \sum_{i=0}^\infty c_i x^i \) is a power series for which each coefficient \( c_i \) is \( 0 \) or \( 1 \). Show that if \( f(2/3) = 3/2 \), then \( f(1/2) \) must be irrational.

Solution:
Suppose by way of contradiction that \( f(1/2) \) is rational. Then \( \sum_{i=0}^\infty c_i 2^{-i} \) is the binary expansion of a rational number, and hence must be eventually periodic; that is, there exist some integers \( m, n \) such that \( c_i = c_{m+i} \) for all \( i \geq n \). We may then write  
\[
f(x) = \sum_{i=0}^{n-1} c_i x^i + \frac{x^n}{1 - x^m} \sum_{i=0}^{m-1} c_{n+i} x^i.
\]  

Evaluating at \( x = 2/3 \), we may equate \( f(2/3) = 3/2 \) with  
\[
\frac{1}{3^{n-1}} \sum_{i=0}^{n-1} c_i 2^i 3^{n-i-1} + \frac{2^n 3^m}{3^{n+m-1} (3^m - 2^m)} \sum_{i=0}^{m-1} c_{n+i} 2^i 3^{m-i-1},
\]  
since all terms on the right-hand side have odd denominator, the same must be true of the sum, a contradiction.  

### Remark
Greg Marks asks whether the assumption that \( f(2/3) = 3/2 \) further ensures that \( f(1/2) \) is transcendental. We do not know of any existing results that would imply this. However, the following result follows from a theorem of T. Tanaka (Algebraic independence of the values of power series generated by linear recurrences, Acta Arith. 74 (1996), 177–190), building upon work of Mahler. Let \( \{a_n\}_{n=0}^\infty \) be a linear recurrent sequence of positive integers with characteristic polynomial \( P \). Suppose that \( P(0), P(1), P(-1) \neq 0 \) and that no two distinct roots of \( P \) have ratio which is a root of unity. Then for \( f(x) = \sum_{n=0}^\infty x^{a_n} \), the values \( f(1/2) \) and \( f(2/3) \) are algebraically independent over \( \mathbb{Q} \). (Note that for \( f \) as in the original problem, the condition on ratios of roots of \( P \) fails.)