For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that:


Since $x^2+bx+c=0$ does not have real solutions, we have $b^2\leq 4c.$
Since $x^2+cx+b=0$ does not have real solutions, we have $c^2\leq 4b.$

Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get \[b^4\leq 16c^2\leq 64b.\]
We apply casework to the value of $b:$

If $b=1,$ then $1\leq 16c^2\leq 64,$ from which $c=1,2.$
If $b=2,$ then $16\leq 16c^2\leq 128,$ from which $c=1,2.$
If $b=3,$ then $81\leq 16c^2\leq 192,$ from which $c=3.$
If $b=4,$ then $256\leq 16c^2\leq 256,$ from which $c=4.$
Together, there are $\boxed{\textbf{(B) } 6}$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$
~MRENTHUSIASM