Gelatinous Polynomials

Consider the polynomial $g(x)=p(x)-x^3$ . We want to find maximum number of integral roots of $g(x)$ subject to the given constraints on the polynomial $p(x)$ . By the fundamental theorem, the polynomial $g(x)$ can be factorized into its roots (assumed to be there are $m$ of them) as follows $g(x)=p(x)-x^3= A \prod_{i=1}^{m} (x-\alpha_i)$ For some real number $A$ and possibly complex numbers $\alpha_i, i=1,2, \ldots, m$ . Since $p(100)=100$ , we require $100-100^3= A\prod_{i=1}^{m}(100-\alpha_i)$ i.e., $A\prod_{i=1}^{m}(100-\alpha_i)=-1\times 2^2\times 3^2 \times 5^2 \times 11 \times 101$ Clearly, there can be at most $m$ integral roots of $g(x)$ which is achieved when $A$ and all of $\alpha_i$ 's are integers. In that case the polynomial $p(x)$ will have all integral coefficients. From the factorization above, we can choose at most $2+2+2+1+1+2=10$ (the last two is for the factor $\pm 1$ ) distinct integral values of the roots $\alpha_i$ , which gives the maximum number of integral solutions for $p(k)=k^3$ .