Saudi Arabian Mathematical Competitions

For a prime $p$ consider the equation
$$p x^{3}+a x^{2}+b x+c=0$$
where $a+b+c=p(p-1)$. Let $x_{1}, x_{2}, x_{3}$ be its roots. From Viète's relation,
$$\begin{gathered} x_{1}+x_{2}+x_{3}=-\frac{a}{p} \\ x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=\frac{b}{p} \\ x_{1} x_{2} x_{3}=-\frac{c}{p} \end{gathered}$$
We have
$$\begin{gathered} \left(x_{1}-1\right)\left(x_{2}-1\right)\left(x_{3}-1\right)=-\frac{c}{p}-\frac{b}{p}-\frac{a}{p}-1 \\ =-\frac{a+b+c}{p}-1=1-p-1=-p \end{gathered}$$
Because $x_{1}, x_{2}, x_{3} \neq 0$, we must have $x_{1}=2, x_{2}=2, x_{3}=1-p$, up to permutation. It follows that
$$5-p=-\frac{a}{p}, \quad 4(2-p)=\frac{b}{p}, \quad 4(1-p)=-\frac{c}{p}$$
hence all triples are $(p(p-5), 4 p(2-p), 4 p(1-p))$.
In our case, $p=2011$ and we obtain
$$(a, b, c)=(2006 \cdot 2011,-4 \cdot 2009 \cdot 2011,4 \cdot 2010 \cdot 2011)$$

As in the previous solution assume that $p$ is a prime and $a+b+1=p(p-1)$. Consider the polynomial
$$P(x)=p x^{3}+a x^{2}+b x+c$$
and let $x_{1}, x_{2}, x_{3} \in \mathbb{Z}^{*}$ be its roots. We have
$$P(x)=p\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)$$
hence
$P(1)=p\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)=p+a+b+c=p+p(p-1)=p^{2}$.
It follows
$$\begin{equation*} \left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)=p \tag{1} \end{equation*}$$
Since $x_{1}, x_{2}, x_{3} \neq 0$, we get $1-x_{1}=-1,1-x_{2}=-1,1-x_{3}=p$, hence $x_{1}=x_{2}=2$, and $x_{3}=1-p$, up to a permutation of $x_{1}, x_{2}, x_{3}$. Therefore
$$\begin{aligned} P(x) & =p(x-2)^{2}(x+p-1)=p\left(x^{2}-4 x+4\right)(x+p-1) \\ & =p x^{3}+p(p-5) x^{2}+4 p(2-p) x+4 p(1-p), \end{aligned}$$
hence $(a, b, c)=(p(p-5), 4 p(2-p), 4 p(1-p))$.