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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?

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  • $\begingroup$ My answer concluded that the 2011 and 2012 does not matter and it could be anything as long as they are t and t+1. Is the consensus that that is true? $\endgroup$ – kaine Sep 12 '13 at 20:16
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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 $c^2 + rx + s = ab$.

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$$a=t^2+rt+s$$ $$b=(t+1)^2+r(t+1)+s=t^2+2t+1+rt+r+s$$ $$ab = c^2+cr+s=(t^2+rt+s)(t^2+2t+1+rt+r+s) = a(a+2t+1+r)$$ $$c = 1/2(\sqrt{r^2-4s+4ab}-r)$$ 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 $\sqrt{r^2-4s+4ab}-r = 2(rt+s+t^2+t)$

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  • $\begingroup$ Could you tell me a bit more about how you simplified the fifth equation after substituting a? $\endgroup$ – Yadnarav3 Sep 13 '13 at 3:18
  • $\begingroup$ I assumed that r,2s,and $2t^2$ would be parts of the number that would appear under the exponent in equation 5 due to the presence of r^2, 4s^2, and 4t^4. I then expanded $(r+2s+2t^2+2k)^2$ to solve for k, got bored, and finished it on wolfram alpha. $\endgroup$ – kaine Sep 13 '13 at 15:19
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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.

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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

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