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?

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?

mr.math · Answer

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
$$a^2-(a-2)^2-(a-3)^2+a(a-2)=2011 \implies \large a=253.$$