No. Suppose that $f(x)=\sum_{k=1}^{20} x^{d_k}$. Then $$f(x)^2=\sum_{k=1}^{20} x^{2d_k}+2\sum_{1\leq i<j\leq 20} x^{d_i+d_j}$$It follows that only the terms of degree $2d_k$ can have odd coefficients. Suppose the coefficient of $x^{2d_k}$ is 9. There exists 4 pairs $(i,j)$ with $i<j$ such that $d_i+d_j=2d_k$. For each of these pairs, note that one of the $d_i$ and $d_j$ is greater than $d_k$. Thus there are at least 4 exponents $d_i$ greater than $d_k$, and at least 4 exponents $d_j$ less than $d_k$. Hence $5 \leq k \leq 16$, which means that there are at most 12 such coefficients.