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?
 A: The quadratic equation $c^2+rc+s-ab=0$ has two solutions, namely
$$
c=(2011r+s+2011\cdot 2012),
$$
or
$$
c=-(2012r + s + 2011\cdot 2012)
$$
And indeed, with this choice of $c$ it works.
A: $$a=t^2+rt+s$$
$$b=(t+1)^2+r(t+1)+s=t^2+2t+1+rt+r+s$$
$$ab = c^2+cr+s=(t^2+rt+s)(t^2+2t+1+rt+r+s) = a(a+2t+1+r)$$
$$c = 1/2(\sqrt{r^2-4s+4ab}-r)$$
c is an integer iff $r^2-4s+4ab$ is a perfect square and $\sqrt{r^2-4s+4ab}-r$ is even
$$r^2-4s+4ab = r^2+4a^2+8at+4a+4ar-4s$$
substiting back in a yields
$$r^2-4s+4ab = (2rt+r+2s+2t^2+2t)^2$$
note that the $\sqrt{r^2-4s+4ab}-r = 2(rt+s+t^2+t)$
A: Subtracting the first equation from the second, we have
$r = b-a-2012^2+2011^2 = b-a - (2012+2011) = (b-2012) + (-a-2011)$ 
Multiplying the first equation by $2012$, the second by $2011$ and subtracting, we have
$s = 2012a - 2011b - 2012\cdot 2011^2 + 2011\cdot 2012^2 = 2012a - 2011b + 2011\cdot 2012$  
So $s - ab = -ab + 2012a - 2011b + 2011\cdot 2012 = (b-2012)\cdot(-a-2011)$
Thus we have clear choices for the sum and product of the roots of $c^2 + rx + s = ab$.
A: 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
