# Covering all but finitely many integers via some given polynomials

Let f be a polynomial with integer coefficients. Let B(f) be the set of all values of f on positive integers. B(f) = {f(n)| n is a positive integer} = {f(1), f(2), ...}

A positive integer k is called "good" if it is a sum of distinct members of B(f). Otherwise we say k is "bad".

For instance, if f(x)=x^3, then it is known that there are finitely many bad numbers. In other words, all but finitely many natural numbers can be written as sum of distinct cubes.

If f(x)=2x^2 + 2, then there are infinitely many bad numbers since every odd number is bad.

My question : if gcd of coefficients of f is 1, are there only finitely many bad numbers?

• $f(n) = (n+1)n = n^2 + n$ is always even. Mar 24 at 13:38

With the right adaptations, the answer should be yes. In particular, the trivially necessary assumption is that the gcd of the values (not just of the coefficients) is 1. Then it is shown in K. F. Roth, G. Szekeres, Some asymptotic formulae in the theory of partitions (1954) that every sufficiently large integer is a sum of distinct values of $$f$$.