Contest math problem algebra proof Let $r, s$ be integers and let
$$a = (2011)^2 + (2011)r + s$$ and 
$$b = (2012)^2 + (2012)r + s$$
Show that there exists an integer $c$ with $c^2 + rc + s = ab$.
Can anyone help me with this?

Choose r = 2 and s = 1
Then $2011^2 + 2.2011 + 1 = a$
and  $2012^2 + 2.2012 + 1 = b$
$a = (2011 + 1)^2$
$b = (2012 + 1)^2$
$ab = (2012.2013)^2$
Let 2012 = u
$ab = (u(u+1))^2$
$ab = u^4 + 2u^3 + u^2$
Let $c = au^2+bu+l$
$ab = (au^2+bu+l)^2 + (au^2 + bu +l)*2 + 1$
$a^2u^4 + b^2u^2+l^2+2(au^2bu+2bul+alu^2)+2au^2+2bu+2c+1$
$a^2u^4+2abu^3+u^2(b^2+2al+2a)+(2bl+2b)u+l^2+2l+1$
Expanding the expression
$a^2 = 1 > a = 1$
$2ab = 2 > b = 1$
$b^2 + 2al + 2a = 1 > l = -1$
Thus $c = (u^2 + u -1)$ which is an integer as u is an integer
Goes to prove that $ab = c^2 + c*2 + 1$ holds true for an integer c.
Thanks
Satish
