Johan Gunardi wrote: Let $ a$ and $ b$ be integers such that the equation $ x^3 - ax^2 - b = 0$ has three integer roots. Prove that $ b = dk^2$, where $ d$ and $ k$ are integers and $ d$ divides $ a$. Let $ u,v,w$ be the three integer roots. We get : $ (1)$ : $ u+v+w=a$ $ (2)$ : $ uv+uw+vw=0$ $ (3)$ : $ uvw=b$ If $ b=0$, the required property is true since $ b=0=1\cdot 0^2$ with $ 1|a$ so we'll now consider $ b\ne 0$ and so $ u,v,w\ne 0$ Let $ u=u'c$ and $ v=v'c$ with $ \gcd(u,v)=c>0$ $ (2)$ $ \implies$ $ u'v'c^2+w(u'+v')c=0$ $ \implies$ $ u'v'c+w(u'+v')=0$ $ \implies$ $ u'+v'|c$ (since $ \gcd(u'v',u'+v')=1$ So $ c=(u'+v')k$ and $ u'v'c+w(u'+v')=0$ becomes $ w=-u'v'k$. So : $ u=u'(u'+v')k$ $ v=v'(u'+v')k$ $ w=-u'v'k$ Then $ (3)$ $ \implies$ $ b=-(u'v'(u'+v')k)^2k$ and $ (1)$ $ \implies$ $ k|a$ Hence the result.