Say I have probabilities $p_i$ and $q_i$, $i=1,\dots, n$ and the following two equations \begin{align}\sum _{i=1}^{n}(-p_{i}^{4} +2p_{i}^{3} -2p_{i}^{2} +p_{i} )&=\sum _{i=1}^{n}(-q_{i}^{4} +2q_{i}^{3} -2q_{i}^{2} +q_{i} )\\[0.2cm]\sum _{i=1}^{n}2(-p_{i}^{4} +2p_{i}^{3} -2p_{i}^{2} +p_{i} )&=\sum _{i=1}^{n}(2p_{i} q_{i}^{2} +2q_{i} p_{i}^{2} -2p_{i}^{2} q_{i}^{2} -4p_{i} q_{i} +p_{i}+q_{i} )\end{align}
Could something (simpler) be said of the relation of $p_i$ to $q_i$? For instance, is it true that $p_i=q_i$ for all $i$?
[I get these equations from equality of means of some distributions I have.]
We substitute, $$x_i = p_i(1-p_i)$$ $$y_i=q_i(1-q_i)$$ Then, from the 1st equation in the above question, $$\sum x_i(1-x_i) = \sum y_i(1-y_i)$$ Such that, $$\sum [x_i(1-x_i) + y_i(1-y_i)] = \sum 2x_i(1-x_i)$$ and using the 2nd equation, $$= \sum [2p_i q_i (q_i + p_i - p_iq_i - 1) + p_i + q_i - 2p_iq_i]$$ $$= \sum [-2p_i q_i (1-p_i)(1-q_i) + (p_i-p_i^2) + (q_i-q_i^2) + (p_i^2 + q_i^2- 2p_iq_i)]$$ $$= \sum [-2 x_i y_i + x_i +y_i + (p_i-q_i)^2]$$ This implies that, $$ \sum [(x_i-y_i)^2 + (p_i-q_i)^2] = 0. $$ Which occurs only if $p_i = q_i$ for all $i=1,\ldots,n$.