In what relationship must the coefficients p and q stand

In what relationship must the coefficients $p$ and $q$ stand so that the $x$-axis is tangent to the curve $y = x^{3}+px+q$?

Step 1: The two conditions

The line $y = 0$ touches the curve at some $x_{0}$ exactly when the point lies on the curve and the slope there is zero: \[x_{0}^{3}+px_{0}+q = 0, \tag{1}\]

\[ y' = 3x^{2}+p \;\Longrightarrow\; 3x_{0}^{2}+p = 0 \;\Longleftrightarrow\; x_{0}^{2} = -\frac{p}{3}. \tag{2} \]

Equation (2) already forces $p\le0$.

Step 2: Eliminate $x_{0}$

From (2), $x_{0}^{3} = x_{0}\cdot x_{0}^{2} = -\dfrac{px_{0}}{3}$; substituting into (1): \[-\frac{px_{0}}{3}+px_{0}+q = \frac{2px_{0}}{3}+q = 0 \;\Longrightarrow\; x_{0} = -\frac{3q}{2p} \qquad \bigl(p\neq0\bigr). \tag{3}\]

Squaring (3) and comparing with (2): \[\frac{9q^{2}}{4p^{2}} = -\frac{p}{3} \;\Longrightarrow\; 27q^{2} = -4p^{3} \;\Longrightarrow\; 4p^{3}+27q^{2} = 0. \tag{4}\]

Step 3: The case $p = 0$

Then (2) gives $x_{0} = 0$ and (1) gives $q = 0$; the curve is $y = x^{3}$, whose tangent at the origin is indeed the $x$-axis. The pair $p = q = 0$ satisfies (4), so no case is lost.

Step 4: Verification

\[p = -3,\ q = 2:\ 4(-27)+27\cdot4 = 0;\quad y = x^{3}-3x+2 = \bigl(x-1\bigr)^{2}\bigl(x+2\bigr),\ \text{touching at } x_{0} = 1 \;\;\checkmark\]

\[p = -3,\ q = -2:\ y = x^{3}-3x-2 = \bigl(x+1\bigr)^{2}\bigl(x-2\bigr),\ x_{0} = -1,\ y'(-1) = 0 \;\;\checkmark\]

\[p = -12,\ q = 16:\ 4(-1728)+27\cdot256 = 0;\quad y = \bigl(x-2\bigr)^{2}\bigl(x+4\bigr) \;\;\checkmark\]