Solve the system of equations in $$x$$, $$y$$, $$x^2 + xy = p^2 + pq$$, $$y^2 + xy = p^2 - pq$$, where $$p$$, $$q$$ are real numbers, $$p \ne 0$$.

PLACEHOLDER ANSWER - the answer is a pair of expressions. Adding the equations gives $$(x + y)^2 = 2p^2$$, so $$x + y = \pm\sqrt{2}\,p$$. Subtracting gives $$(x + y)(x - y) = 2pq$$, so $$x - y = \pm\sqrt{2}\,q$$ with the same sign. Hence $$(x, y) = \left(\frac{p + q}{\sqrt{2}}, \frac{p - q}{\sqrt{2}}\right)$$ or $$\left(-\frac{p + q}{\sqrt{2}}, -\frac{p - q}{\sqrt{2}}\right)$$.