Sum of polynomial coefficient 
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x)=2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?"
  After some calculations, Jon says, "There is more than one such polynomial."
  Steve says, "You’re right. Here is the value of $a$." He writes down a positive integer and asks, "Can you tell me the value of $c$?"
  Jon says, "There are still two possible values of $c$."
  Find the sum of the two possible values of $c$.

$$P(x) = 2x^3-2ax^2+(a-9)(a+9)x-c$$
Let the roots be $r_1, r_2, r_3$ then:
$$P(x) = 2(x - r_1)(x - r_2)(x - r_3)$$
$$= 2x^3  - 2x^2\overbrace{(r_1 + r_2 + r_3)}^{=a} + x\overbrace{(2r_1r_2 + 2r_1r_3 + 2r_2r_3)}^{= (a-9)(a+9)} - \overbrace{2r_1r_2r_3}^{=c}$$
Yikes!
$$r_1 + r_2 + r_3 = a$$
$$2(r_1r_2 + r_1r_3 + r_2r_3) = (a-9)(a+9)$$
 A: We have
$$2(r_1r_2+r_2r_3+r_3r_1) = a^2 - 81 = (r_1+r_2+r_3)^2 - 81$$
This gives us
$$r_1^2 + r_2^2 + r_3^2 = 81$$
Since $81 \equiv 1\pmod8$, we obtain that one of them is odd, say $r_1$ and the other two ($r_2 = 2k_2$ and $r_3 = 2k_3$) are even. Further, $1 \leq r_1,r_2,r_3 \leq 8$, i.e., $1 \leq k_2,k_3 \leq 4$ and $r_1 \in \{1,3,5,7\}$.


*

*$r_1 = 1$. This gives us $k_2^2 + k_3^2 = 20$. The only possible solution is $(4,2)$ and $(2,4)$.

*$r_2 = 3$. This gives us $k_2^2 + k_3^2 = 18$. The only possible solution is $(3,3)$ and $(3,3)$.

*$r_2 = 5$. This gives us $k_2^2 + k_3^2 = 14$. No solution exists.

*$r_2 = 7$. This gives us $k_2^2 + k_3^2 = 8$. This gives $k_2=k_3=2$.


Hence, the possible roots are $(1,4,8)$, $(3,6,6)$ and $(7,4,4)$. Note that the value of $r_1+r_2+r_3$ are $13$, $15$ and $15$ respectively. Further given $a$, there are two possible values of $c$. This means the roots are $(3,6,6)$ or $(7,4,4)$. Hence, the possible values of $c$ are $216$ and $224$.
A: You could continue this way:
$$a^2-2(r_1r_2+r_1r_3+r_2r_3)=r_1^2+r_2^2+r_3^2=81$$
