$x_1+x_2=b$ and $x_1x_2=c\Longrightarrow{}5=x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=b^2-2c$. Obviously, $b$ is odd $\Longrightarrow{}b=2k-1\Longrightarrow{}5+2c=b^2=(2k-1)^2=4k^2-4k+1\Longrightarrow{}c=2k^2-2k-2$. So answer is $(b,c)=(2k-1,2k^2-2k-2)$ $\forall{k}\in{N}$

from Vieta's formulas x1+x2 =b and (x1+x2)^2 =X1^2+X2^2 +2x1x2. 2x1x2=c so x1^2 +x2^2=5=b^2-c

Do not forget b^2 -4c >0 ==> finite value for k