2011 AIME I Problems/Problem 15

Problem

For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ , $b$ , and $c$ . Find $|a| + |b| + |c|$ .

With Vieta's formula, we know that $a+b+c = 0$ , and $ab+bc+ac = -2011$ .

$a,b,c\neq 0$ since any one being zero will make the the other 2 $\pm sqrt{2011}$ .

$a = -(b+c)$ . WLOG, let $|a| \ge |b| \ge |c|$ .

Then if $a > 0$ , then $b,c < 0$ and if $a < 0$ , $b,c > 0$ .

$ab+bc+ac = -2011 = a(b+c)+bc = -a^2+bc$

$a^2 = 2011 + bc$

We know that $b$ , $c$ have the same sign. So $|a| \ge 45$ . ( $44^2<2011$ and $45^2 = 2025$ )

Also, $bc$ maximize when $b = c$ if we fixed $a$ . Hence, $2011 = a^2 - bc < \frac{3}{4}a^2$ .

So $a ^2 < \frac{(4)2011}{3} = 2681+\frac{1}{3}$ .

$52^2 = 2704$ so $|a| \le 51$ .

Now we have limited a to $45\le |a| \le 51$ .

Let's us analyze $a^2 = 2011 + bc$ .

Here is a table:

We can tell we don't need to bother with $45$ ,

$105 = (3)(5)(7)$ , So $46$ won't work. $198/47 > 4$ ,

$198 is not divisible by 5$ , $198/6 = 33$ , which is too small to get $47$

$293/48 > 6$ , $293$ is not divisible by $7$ or $8$ or $9$ , we can clearly tell that $10$ is too much

Hence, $a = 49$ , $a^2 -2011 = 390$ . $b = 39$ , $c = 10$ .

Answer: $098$