# Prime numbers $(x,c,p)$ such that $x^3-p x^2-cx-5c=0$

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 $p=3$.
can you please explain why $25$ divides $5c$ ? I can only see $25$ divides $10c$ –  Belgi Aug 7 '12 at 3:20
@Belgi: Thanks. I should have written $25$ divides $10c$. Fixed. It still follows that $c=5$. –  André Nicolas Aug 7 '12 at 3:23