Let the roots to the Cris equation be $\alpha$ and $\beta$. Then the roots to Cristian's equation are $\alpha^2$ and $\beta^2$ By Vieta, we can see that \begin{align*} \alpha+\beta&=\frac{b}{2} \\ \alpha \beta &= -\frac{c}{2} \\ \alpha^2+\beta^2&=b+1 \\ \alpha^2 \beta^2 &=-(c+1) \end{align*}Thus, $-(c+1)=\left( -\frac{c}{2} \right)^2 \implies c^2+4c+4=0 \implies c=-2$ Also, $(b+1)+2\left( -\frac{c}{2} \right)=\left( \frac{b}{2} \right)^2 \implies b^2-4b-12=0 \implies b=-2$ or $b=6$. Thus, $(b,c)=(-2,-2)$ or $(-2,6)$