Mathematical Reflections 2012, Issue 1 - Problem J217

Problem:
If $a, b, c$ are integers such that $a^2 + 2bc = 1$ and $b^2 + 2ca = 2012$, find all possible values
of $c^2 + 2ab$.

Proposed by Titu Andreescu.

Solution:

Subtracting the two given equations, we have $(b-a)(b+a-2c)=2011$, so we get $$\left\{ \begin{array}{lll} b & = & a \pm 1 \\ 2c & = & b+a \mp 2011 \end{array} \right. \qquad \left\{ \begin{array}{lll} b & = & a \pm 2011 \\ 2c & = & b+a \mp 1. \end{array} \right.$$
Substituting these values into the equation $a^2+2bc=1$, we obtain the four equations

(i) $3a^2-2008a-2011=0$
(ii) $3a^2+2008a-2011=0$
(iii) $3a^2+6032a+2010\cdot2011-1=0$
(iv) $3a^2-6032a+2010\cdot2011-1=0$.

Equation (i) gives $(a+1)(3a-2011)=0$, so $a=-1, b=0, c=-1006$.
Equation (ii) gives $(a-1)(3a+2011)=0$, so $a=1, b=0, c=1006$.
Equations (iii) and (iv) have no solution since their discriminant is negative.
In conclusion, $c^2+2ab=1006^2=1012036$.