We will prove by induction that we can choose $a_{i}$ such that $(...((x^{2}+a_{1})^{2}+...)^{2}+a_{i})^{2}$ has only $n-i$ possible remainders modulo $2n-1$. For $i=1$, we have to show that an $a_{1}$ exists such that $(x^{2}+a_{1})^{2}$ has only $n-1$ possible remainders modulo $2n-1$. Note that $x^{2}$ has at most $n$ possible remainders modulo $2n-1$. Let $b,c$ be two of them. Let $a_{1}\equiv \frac{b+c}{2}$ (mod $2n-1$). Then we get $b+a_{1}\equiv -(c+a_{1})$ (mod $2n-1$) and so $(b+a_{1})^{2}\equiv (c+a_{1}^{2})$ (mod $2n-1$). Thus $(x^{2}+a_{1})^{2}$ has at most $n-2+1=n-1$ possible remainders. The inductive step works in the same way. Therefore we have found $a_{1},...,a_{n-1}$ such that $(... (((x^2 + a_1)^2 + a_2)^2 + ...)^2 + a_{n-1})^2$ has only one possible remainder modulo $2n-1$ and hence we can find an $a_{n}$ such that $(... (((x^2 + a_1)^2 + a_2)^2 + ...)^2 + a_{n-1})^2 +a_{n}$ is divisible by $2n-1$.