Finding possible values of $|a| + |b| + |c|$ for solutions of $x^3 - 2019x + m = 0$

Question

There is an integer $m$ in the polynomial $x^3 - 2019x + m = 0$ so that the polynomial has three solutions namely $a$, $b$, and $c$, where $a$ and $b$ are positive integers and $c$ is an integer only (can be positive, negative or 0). Determine the possible values of $|a| + |b| + |c|$.

A. 2018

B. 100

C. 0

D. 55

E. 201

My approach:

Since $\displaystyle a$ , $\displaystyle b$ , and $\displaystyle c$ are the three solutions of the polynomial $\displaystyle x^{3} -2019x+m\ =\ 0$, we can use Vieta's formulas to find the sum and product of the roots.

Vieta's formulas state that for a cubic polynomial of the form $\displaystyle ax^{3} +bx^{2} +cx+d\ =\ 0$, the sum of the roots is given by $\displaystyle -\frac{ b}{ a}$ and the product of the roots is given by $\displaystyle -\frac{ d}{ a}$.

In this case, we have a cubic polynomial of the form $\displaystyle x^{3} -2019x+m\ =\ 0$, so the sum of the roots is given by $\displaystyle -\frac{ -2019}{ 1}\ =\ 2019$ and the product of the roots is given by $\displaystyle -\frac{ m}{ 1}\ =\ -m$.

Since we know that $\displaystyle a$ and $\displaystyle b$ are positive integers and $\displaystyle c$ is an integer, we can use these formulas to find possible values for $\displaystyle |a|+|b|+|c|$.

Since the sum of the roots is 2019, we have that $\displaystyle a+b+c=2019$. Since $\displaystyle a$ and $\displaystyle b$ are positive integers, their sum must be an integer greater than or equal to 2. Therefore, we have that $\displaystyle c=2019-(a+b)$ must be an integer greater than or equal to 2017.

Since $\displaystyle c$ is an integer greater than or equal to 2017, its absolute value must also be greater than or equal to 2017. Therefore, we have that $\displaystyle |a|+|b|+|c|\geq |c|\geq 2017$. Looking at the answer choices, we see that only choice A (2018) satisfies this condition. Therefore, the possible value for $\displaystyle |a|+|b|+|c|$ is 2018.

Is this correct?

Answer 1

This is not an answer. Done for the fun of using algebra.

Using brute force, the discriminant of the cubic equation is $$\Delta=27 \left(1219284868-m^2\right)$$ So, for three real roots, $|m|<1346 \sqrt{673}$

Using the trigonometric solution fot three real roots, the solutions are $$x_k=2 \sqrt{673} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(-\frac{m}{1346 \sqrt{673}}\right)\right)\right)\qquad \text{for} \qquad k=0,1,2$$

To have $x_1>0$, we need $0 <m<1346 \sqrt{673}$ but, over this range $x_2<0$ and $x_3>0$ and $$\sqrt{2019}\leq x_1+x_3\leq 2 \sqrt{673}$$ Therefore $$2 \sqrt{2019} \leq x_1+x_3+|x_2| \leq 4 \sqrt{673}$$

Thsi does not leave much room for integer numbers since $\lceil 2 \sqrt{2019} \rceil =90$ and $\lfloor 4 \sqrt{673}\rfloor=103$.

Take into account the fact that, up to the last line, I still did not specify that $(x_1,x_2,x_3)$ are integer numbers.

Beside the one given by @Sathvik (which corresponds to $m=24050$), there is no other solution.

Answer 2

No, you have made a few errors.

Since $a,b,c$ are roots of the given cubic, $$a+b+c=0\implies c=-(a+b).$$ Also, $$ab+bc+ca=-2019\implies a^2+ab+b^2=2019.$$ Note, $a,b>0$ and $c=-(a+b)\implies |a|+|b|+|c|=2(a+b).$
Now, you can eliminate all options other than $100$. Also, $a=13, b=37, c=-50$ is an example where all given conditions are satisfied.