Question

AMC 10 2012 #24
Let a, b, and c be positive integers with \[a \ge b \ge c\] such that \[a^2-b^2-c^2+ab=2011\] and \[a^2+3b^2+3c^2-3ab-2ac-2bc=-1997\]. What is a?

Answer 1

We can write the second equation as
\[(a-c)^2+(a-b)^2+(b-c)^2+b^2+c^2-a^2-ab=-1997.\]
But \(b^2+c^2-a^2-ab=-2011\), and that gives
\((a-c)^2+(a-b)^2+(b-c)^2=14=3^2+2^2+1^2.\) That's
\(a-c=3 \implies c=a-3\), \(a-b=2 \implies b=a-2\) and \(b-c=1\).
Plug this into the first equation and solve the quadratic equation
\[a^2-(a-2)^2-(a-3)^2+a(a-2)=2011 \implies \large a=253.\]