Saudi Arabia Mathematical Competitions

For a prime $p$ consider the equation $$p{x}^3+a{x}^2+bx+c=0$$ where $a+b+c=p(p-1)$. Let $x_1,x_2,x_3$ be its roots. From Viète's relation, $$\begin{array}{c}{x}_1+x_2+x_3=-\frac{a}{p}\\ x_1{x}_2+x_2{x}_3+x_3{x}_1=\frac{b}{p}\\ x_1{x}_2{x}_3=-\frac{c}{p}\end{array}$$ We have $$\begin{array}{c}\left(x_1-1\right)\left(x_2-1\right)\left(x_3-1\right)=-\frac{c}{p}-\frac{b}{p}-\frac{a}{p}-1\\ =-\frac{a+b+c}{p}-1=1-p-1=-p\end{array}$$ Because $x_1,x_2,x_3\ne 0$, we must have $x_1=2,x_2=2,x_3=1-p$, up to permutation. It follows that $$5-p=-\frac{a}{p},4(2-p)=\frac{b}{p},4(1-p)=-\frac{c}{p}$$ hence all triples are $(p(p-5),4p(2-p),4p(1-p))$. In our case, $p=2011$ and we obtain $$(a,b,c)=(2006\cdot 2011,-4\cdot 2009\cdot 2011,4\cdot 2010\cdot 2011)$$

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Alternative solution.

As in the previous solution assume that $p$ is a prime and $a+b+1=p(p-1)$. Consider the polynomial $$P(x)=p{x}^3+a{x}^2+bx+c$$ and let $x_1,x_2,x_3\in {\mathbb{Z}}^{\ast }$ 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 $$\begin{array}{c}\left(1-x_1\right)\left(1-x_2\right)\left(1-x_3\right)=p\end{array}\text{\hspace{1em}(1)}$$ Since $x_1,x_2,x_3\ne 0$, 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 $$\begin{array}{rl}P(x)& =p(x-2{)}^2(x+p-1)=p\left(x^2-4x+4\right)(x+p-1)\\ & =p{x}^3+p(p-5)x^2+4p(2-p)x+4p(1-p),\end{array}$$ hence $(a,b,c)=(p(p-5),4p(2-p),4p(1-p))$.