# 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 $|a| + |b| + |c|$.

## Solution 1

From Vieta's formulas, we know that $a+b+c = 0$, and $ab+bc+ac = -2011$. Thus $a = -(b+c)$. All three of $a$, $b$, and $c$ are non-zero: say, if $a=0$, then $b=-c=\pm\sqrt{2011}$ (which is not an integer). $\textsc{wlog}$, let $|a| \ge |b| \ge |c|$. If $a > 0$, then $b,c < 0$ and if $a < 0$, then $b,c > 0$. We have $$-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc$$ Thus $a^2 = 2011 + bc$. We know that $b$, $c$ have the same sign. So $|a| \ge 45 = \lceil \sqrt{2011} \rceil$.

Also, if we fix $a$, $b+c$ is fixed, so $bc$ is maximized when $b = c$ . Hence, $$2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 < \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}$$ So $|a| \le 51$. Thus we have bounded $a$ as $45\le |a| \le 51$, i.e. $45\le |b+c| \le 51$ since $a=-(b+c)$. Let's analyze $bc=(b+c)^2-2011$. Here is a table:

$|a|$$bc=2011-a^2$
$45$$14$
$46$$105$
$47$$198$
$48$$293$
$49$$390$

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: $\boxed{098}$

## Solution 2

Starting off like the previous solution, we know that $a + b + c = 0$, and $ab + bc + ac = -2011$.

Therefore, $c = -b-a$.

Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$.

Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$.

Therefore, a sum ($a+b$) squared minus a product ($ab$) gives $2011$..

We can guess and check different $a+b$’s starting with $45$ since $44^2 < 2011$.

$45^2 = 2025$ therefore $ab = 2025-2011 = 14$.

Since no factors of $14$ can sum to $45$ ($1+14$ being the largest sum), a + b cannot equal $45$.

$46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$.

$5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either.

We can continue to do this until we reach $49$.

$49^2 = 2401$ making $ab = 390 = 2 * 3 * 5* 13$.

$3 * 13 + 2* 5 = 49$, so one root is $10$ and another is $39$. The roots sum to zero, so the last root must be $-49$.

$|-49|+10+39 = \boxed{098}$.

## Solution 3

Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$. Now, due to the first equation, let us say that $a+b=-c$, meaning that $a,b>0$ and $c<0$. Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$, respectively. Since $c$ is less than 0, it equals $-a-b$. Therefore, $|c|=|-a-b|=a+b$, meaning $|a|+|b|+|c|=2(a+b)$. We now apply Newton's sums to get that $a^2+b^2+ab=2011$,or $(a+b)^2-ab=2011$. Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$, so $2(a+b)=\boxed{098}$.

## Solution 4

We have

$(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc$


As a result, we have

$a+b+c=0$

$ab+bc+ac=-2011$

$abc=-m$

So, $a=-b-c$

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

Solve $b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}$ and $\Delta =8044-3c^2=k^2$, where $k$ is an integer

Cause $89<\sqrt{8044}<90$

So, after we tried for $2$ times, we get $k=88$ and $c=10$

then $b=39$, $a=-b-c=-49$

As a result, $|a|+|b|+|c|=10+39+49=\boxed{098}$

## Solution 5 (mod to help bash)

First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\implies b^2+bc+c^2=2011$. Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\equiv \{1,2\} \pmod{3}$, and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$, testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$, $c=39, -49$. Choosing $c$ positive we get $a=-49$, so $|a|+|b|+|c|=10+29+39=\boxed{098}$ ~firebolt360

## Solution 6

Note that $-c=b+a$, so $c^2=a^2+2ab+b^2$, or $-c^2+ab=-a^2-ab-b^2$. Also, $ab+bc+ca=-2011$, so $(a+b)c+ab=-c^2+ab=-2011$. Substituting $-c^2+ab=-a^2-ab-b^2$, we can obtain $a^2+ab+b^2=2011$, or $\frac{a^3-b^3}{a-b}=2011$. If it is not known that $2011$ is prime, it may be proved in $5$ minutes or so by checking all primes up to $43$. If $2011$ divided either of $a, b$, then in order for $a^3-b^3$ to contain an extra copy of $2011$, both $a, b$ would need to be divisible by $2021$. But then $c$ would also be divisible by $2011$, and the sum $ab+bc+ca$ would clearly be divisible by $2011^2$.

By LTE, $v_{2011}(a^3-b^3)=v_{2011}(a-b)$ if $a-b$ is divisible by $2011$ and neither $a,b$ are divisible by $2011$. Thus, the only possibility remaining is if $a-b$ did not divide $2011$. Let $a=k+b$. Then, we have $(b+k)^3-b^3=2011k$. Rearranging gives $3b(b+k)=2011-k^2$. As in the above solutions, we may eliminate certain values of $k$ by using mods. Then, we may test values until we obtain $k=29$, and $a=10$. Thus, $b=39$, $c=-49$, and our answer is $49+39+10=098$.