What relation this problem has to the number theory?

@nnosipov, the fact that You know one (or more) solution (algebra-based) to this task, does not mean that he must expect such solution. I also know Your solution, but Im willing to see more different solutions.

Let $m$ be a real solution of $x^3+px+q=0$. Then, consider the quadratic equation $mx^2+px+q=0$ whose one solution is $m$ which is real. Since one of the root is real, the discriminant must be non negative and hence $p^2\ge 4qm$ as required.

How can we invoke this without replacing all $x$ by $m$.I think we can't do this unless replacing all $x$ by $m,$ and we need to use the formula for cubic equation here.

mathmdmb wrote: How can we invoke this without replacing all $x$ by $m$.I think we can't do this unless replacing all $x$ by $m,$ and we need to use the formula for cubic equation here. What are you asking for ? gouthamphilomath is perfectly right (and btw congrats for the quite nice solution !) $m$ root of cubic $x^3+px+q=0$ $\iff$ $m^3+pm+q=0$ $m$ root of quadratic $mx^2+px+q=0$ $\iff$ $m^3+pm+q=0$ So $m$ root of cubic $x^3+px+q=0$ $\iff$ $m$ root of quadratic $mx^2+px+q=0$ $\implies$ quadratic $mx^2+px+q=0$ has real root hence the result. You can write exactly the same thing in another way : $x^3+px+q=0$ $\implies$ $4x^4+4px^2+4qx=0$ $\implies$ $(2x^2+p)^2=p^2-4qx$ hence the result

pco wrote: mathmdmb wrote: How can we invoke this without replacing all $x$ by $m$.I think we can't do this unless replacing all $x$ by $m,$ and we need to use the formula for cubic equation here. What are you asking for ? gouthamphilomath is perfectly right (and btw congrats for the quite nice solution !) $m$ root of cubic $x^3+px+q=0$ $\iff$ $m^3+pm+q=0$ $m$ root of quadratic $mx^2+px+q=0$ $\iff$ $m^3+pm+q=0$ So $m$ root of cubic $x^3+px+q=0$ $\iff$ $m$ root of quadratic $mx^2+px+q=0$ $\implies$ quadratic $mx^2+px+q=0$ has real root hence the result. You can write exactly the same thing in another way : $x^3+px+q=0$ $\implies$ $4x^4+4px^2+4qx=0$ $\implies$ $(2x^2+p)^2=p^2-4qx$ hence the result yes that point.I wasn't clear whether we can really do this or not.Thanks a lot.