smc

A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that: 1. Since $x^2+bx+c=0$ does not have real solutions, we have $b^2\le 4c.$ 2. Since $x^2+cx+b=0$ does not have real solutions, we have $c^2\le 4b.$ Squaring the first inequality, we get $b^4\le 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\le 64b.$ Combining these results, we get $$b^4\le 16c^2\le 64b.$$ We apply casework to the value of $b:$ If $b=1,$ then $1\le 16c^2\le 64,$ from which $c=1,2.$ If $b=2,$ then $16\le 16c^2\le 128,$ from which $c=1,2.$ If $b=3,$ then $81\le 16c^2\le 192,$ from which $c=3.$ If $b=4,$ then $256\le 16c^2\le 256,$ from which $c=4.$ Together, there are $\boxed{6}$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$