Let's (x

_{1}, x_{2}) and (x_{2}, y_{2}) are solutions to the systemx

y-1=x

^{2}+ y^{2}= 17y-1=x

This means that

x

y

_{1}^{2}+y_{1}^{2}= 17 (1a)y

_{1}-x_{1}=1 (1b)and

x

y

_{2}^{2}+y_{2}^{2}= 17 (2a)y

_{2}-x_{2}=1 (2b)Substructing (2a) from (1a) and from (2b) from (1b) results

(x

_{1}-x_{2})(x_{1}+x_{2})+(y_{1}-y_{2})(y_{1}+y_{2})=0 and y_{1}-y_{2}=x_{1}-x_{2}.Since x

_{1}≠x_{2}, substituting the second equation into the first one gives(x

_{1}-x_{2})[(x_{1}+x_{2})+(y_{1}+y_{2})]=0, from which follows that x_{1}+x_{2}+y_{1}+y_{2}=0.