Ireland_2017

Let $y=1-x$. Then $x(2-x)=(1-y)(1+y)=1-y^2$, and so an equivalent version of the equation is $(1-y^2{)}^2=1+2y^2$, i.e., $y^4=4y^2$. Thus $y=0,\pm 2$ and $x=-1,1,3$.

Alternatively, the given equation could be written as $x^2(2-x{)}^2-1=2(1-x{)}^2$. Using the difference of two squares, this becomes $$\begin{array}{rl}(x(2-x)-1)(x(2-x)+1)& =2(1-x{)}^2,\text{ i.e.}\\ (1-x{)}^2(x^2-2x-1)& =2(1-x{)}^2.\end{array}$$ Hence, one solution is $x=1$ and the others are the roots of $x^2-2x-3=(x+1)(x-3)$, that is $x=-1$ and $x=3$.