\section*{Problem 12}

Find all real \(\mathrm{p}, \mathrm{q}, \mathrm{a}, \mathrm{b}\) such that we have \((2 \mathrm{x}-1)^{20}-(\mathrm{ax}+\mathrm{b})^{20}=\left(\mathrm{x}^{2}+\mathrm{px}+\mathrm{q}\right)^{10}\) for all \(\mathrm{x}\).


\section*{Solution}

Comparing coefficients of \(\mathrm{x}^{20}\), we must have \(\mathrm{a}=\left(2^{20}-1\right)^{1 / 20}\) (note that we allow either the positive or the negative root).

Set \(x=1 / 2\). Then we must have \((a x+b)^{20}=0=\left(x^{2}+p x+q\right)^{10}\), and hence \(a x+b=0\) and \(x^{2}+p x+q\) \(=0\). So \(b=-a / 2\), and \(1 / 4+p / 2+q=0\).

Set \(x=0\). Then we get \(q^{10}=1-b^{20}=1 / 2^{20}\), so \(q=1 / 4\) or \(-1 / 4\), and \(p=-1\) or 0 respectively. Comparing the coefficients of \(x^{19}\), we must have \(p=-1\) and \(q=1 / 4\). So, if there is a solution, then it must be: \(\mathrm{a}=\left(2^{20}-1\right)^{1 / 20}, \mathrm{~b}=-\mathrm{a} / 2, \mathrm{p}=-1, \mathrm{q}=1 / 4\). This is indeed a solution because with these values, the lhs \(=2^{20}(x-1 / 2)^{20}-(x-1 / 2)^{20} a^{20}=(x-1 / 2)^{20}=\left(x^{2}-x+1 / 4\right)^{10}=\) rhs.

