How should I proceed to find all prime numbers $x,c,p$ such that $$x^3-px^2-cx-5c=0$$
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If $x^3-px^2-cx-5c=0$, then $x$ divides $5c$. Since $x$ and $c$ are prime, we have the two possibilities $x=c$ and $x=5$.
Suppose $x=c$. Substitute. We get $c^3-(p+1)c^2-5c=0$, so $c^2-(p+1)c-5=0$, so $c$ divides $5$, so $c=5$. Therefore definitely $x=5$.
Substitute. We get $125-25p-10c=0$. Since $25$ divides $10c$, it follows that $c=5$. Thus $p=3$.