What is the sum of all possible values of $\gamma$

Question

The polynomial $x^3 +ax^2 +bx+c$ has three roots $\alpha ≤ \beta ≤ \gamma$, all of which are positive integers. Given that $2^2(a) + 2^1(b) + 2^0(c) = −2^{11}$ what is the sum of all possible values of $γ$?

I tried using Vieta's and finding some relations there such as $-a = \alpha+\beta + \gamma$, to get that $2^2(a) + 2^1(b) + 2^0(c) = -4(\alpha+\beta+\gamma)+2(\alpha \beta +\alpha \gamma + \beta \gamma) -\alpha \beta \gamma = -2^{11}$ but that didn't help me.

Answer 1

Let $p(x) = x^3 + ax^2 + bx + c$. Then $$ p(2) = 2^3 + 2^2a+2b+c = 2^3 - 2^{11}. $$ On the other hand, since $p(x) = (x-\alpha)(x-\beta)(x-\gamma)$, it follows that $$ (2-\alpha)(2-\beta)(2-\gamma) = 2^3 - 2^{11}. $$ Since the right-hand side is negative, either one or three of the terms in the left-hand side must be negative.

Suppose only one term is negative. Then this term must be $2-\gamma$, as $2-\gamma\le 2-\beta\le 2-\alpha$. Since $\alpha$ and $\beta$ are positive integers, the only way $2-\alpha>0$ and $2-\beta >0$ is if $\alpha = \beta = 1$. This implies that $2 - \gamma = 2^3 - 2^{11}$, so $\gamma = 2^{11} - 2^3 + 2$ is one possibility.

If all three terms are negative, then take advantage of the fact that all three terms must be negative, and factor the right-hand side. For ease, rewrite the equation as $$ (\alpha-2)(\beta-2)(\gamma-2) = 2^{11}-2^3 = 2040. $$ Note that $\gamma-2\ge\beta-2\ge\alpha-2$, so we are looking for three positive factors of $2040$, with $\gamma-2$ being the greatest of those three. Now $$ 2^{11}-2^3 = 2^3(2^8-1) = 2^3(2^4-1)(2^4+1) = 2^3(2^2-1)(2^2+1)(2^4+1) = 2^3\times 3\times 5\times 17.$$ There are $(3+1)\times(1+1)\times(1+1)\times(1+1) = 32$ factors of $2040$, so it's now just a matter of figuring out which of these factors can be equal to $\gamma-2$ for some $\gamma$. Since $\gamma-2$ has to be the largest factor, we must have that $\gamma-2\ge 17$. Otherwise, one of the other two factors will be divisible by $17$, and hence larger than $\gamma-2$, contradiction. So $\gamma-2$ cannot be some factor less than $17$.

I now claim that $\gamma-2$ can be any factor greater than or equal to $17$. To do this, we construct values of $\alpha$ and $\beta$ to make it work. Let $n = \gamma-2$. Construct $\alpha$ and $\beta$ as follows:

• If $17$ does not divide $n$, then let $\beta - 2 = 17$, and let $\alpha - 2 = \frac{2040}{17n} = \frac{120}{n}$. Since $n\ge 17$, we have $\beta\le\gamma$, and $\alpha-2\le\frac{120}{17}<17 = \beta-2\implies\alpha\le\beta$. We've thus found $(\alpha,\beta,\gamma)$ with $\alpha\le\beta\le\gamma$ and $\gamma -2 = n$, where $n\ge 17$ divides $2040$, but $17$ does not divide $n$, so $\gamma = 2+n$ is a possible value.
• Otherwise, suppose $17$ divides $n$. Note that $\frac{2040}{17} = 120$. It follows that $\frac{2040}{n}$ must divide $120$. Now, we can find two factors of $120$ which multiply to $120$ and are both less than $17$ (namely, $8$ and $15$). Since $\frac{2040}{n}$ divides $120$, we can find two factors less than $17$ which multiply to $\frac{2040}{n}$. Set $\alpha-2$ and $\beta-2$ equal to those two factors (set $\alpha-2$ to be the lower of the two). Then you can easily see that $\alpha\le\beta\le\gamma$, so $\gamma = 2+n$ is a possible value as well.

So, if you take all of the factors of $2040$ greater than or equal to $17$, and add $2$ to them, and add them all up, you get your desired answer. Note that the answer from the first part, when only one term is negative, i.e. $\gamma = 2^{11}-2^3+2$, also comes out of this method, so there is no need to add the two parts.

(If you're looking for an easier way to add all of these factors, note that you are adding $2+$ every factor of $2040$ that is at least $17$. This means that you are not considering the factors $1$, $2$, $3$, $4$, $5$, $6$, $8$, $10$, $12$, and $15$. So if you know a formula for the sum of the divisors of a given number, then you can just apply the sum and subtract appropriately.)