So this is a 9-10th grade, math olympiad problem I found. Define the parabola $y=ax^2+bx+c$ such that $a,b,c$ are positive integers. Suppose that the roots of the quadratic equation $ax^2+bx+c=0$ are $x_1,x_2$ such that $|x_1|,|x_2|>1$, then compute the minimum value of $abc$ and when the minimum occur what is the value of $a+b+c$. Maybe there is a way to solve this using advanced math, calculus, but I am sure there is an elementary way of solving this without using any calculus.
[Math] Quadratic equation, math olympiad question
contest-mathquadraticsreal-analysissystems of equations
Related Solutions
Let $$m^2=n^2+204n$$ $$k^2=n^2+204n+10404=(n+102)^2$$ Then $$(k-m)(k+m)=2^2\cdot 3^2\cdot 17^2$$ Since $k-m$ and $k+m$ have the same parity, they must be even. Write:
$$\frac{k-m}2\frac{k+m}2=3^2\cdot17^2$$
Since $k>m$ there are only two options: $k-m=2$ and $k-m=18$. The former gives $m=2600$ and the latter $m=280$. Thus we are interested in the former, which gives $n=2500$.
Without loss of generality, suppose $A < B < C < D$.
Notice that if $A \ge 3$, then $B \ge 4$, $C \ge 5$, $D \ge 6$, and then $\tfrac{1}{A}+\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} \le \tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{5}+\tfrac{1}{6} = \tfrac{19}{20} < 1.1$. So we need $A = 1$ or $A = 2$.
Case 1: $A = 2$. Then we need $\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{3}{5}$ with $2 < B < C < D$.
Since $\tfrac{3}{B} > \tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{3}{5}$, we must have $B < 5$, i.e. $B = 3$ or $B = 4$.
If $B = 4$, we need $\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{7}{20}$ with $4 < C < D$. Since $\tfrac{2}{C} > \tfrac{1}{C}+\tfrac{1}{D} = \tfrac{7}{20}$, we must have $C < \tfrac{40}{7}$, i.e. $C \le 5$. Since $4 < C \le 5$, we must have $C = 5$, but then $D = \tfrac{20}{3}$, which is not an integer. So there are no solutions with $A = 2$ and $B = 4$.
If $B = 3$, we need $\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{4}{15}$ with $3 < C < D$. Since $\tfrac{2}{C} > \tfrac{1}{C}+\tfrac{1}{D} = \tfrac{4}{15}$, we must have $C < \tfrac{15}{2}$, i.e. $C \le 7$. Testing $C = 4, 5, 6, 7$ yields $D = 60, 15, 10, \tfrac{105}{13}$ respectively. In this case, the smallest sum where $C$ and $D$ are integers is $21$ which occurs for $(A,B,C,D) = (2,3,6,10)$.
Case 2: $A = 1$. Then, we need $\tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{1}{10}$. But since $\tfrac{3}{D} < \tfrac{1}{B}+\tfrac{1}{C}+\tfrac{1}{D} = \tfrac{1}{10}$, any solution in this case will have $D > 30$, and thus, $A+B+C+D > 30 > 21$. So we will not find a smaller sum in this case.
Therefore, the minimum sum is $21$.
Note that if you just need to get an answer quickly without a rigorous proof, then you can probably just guess and check until you find something reasonably small. In problems with Egyptian fractions (fractions with numerator $1$), the sum $\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{6} = 1$ comes up a lot, namely it is the smallest set of distinct Egyptian fractions that add up to $1$. So it's not too hard to build off of that to get $\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{6}+\tfrac{1}{10} = \tfrac{11}{10}$. I'm not sure if there is an easy way to convince yourself that's the smallest sum though.
Best Answer
$f(x) = ax^2+bx+c$ cannot have any positive root when $a$, $b$, $c$ are all positive. So we have $x_1, x_2<-1$.
The midpoint between the roots is $-b/2a$. That must be less than $-1$, so we have $$ \tag{1} \frac{-b}{2a}<-1 \implies \frac{b}{2a}>1 \implies b>2a $$ Furthermore the larger root is less than $-1$ too, so $$ \tag{2} f(-1) >0 \implies a-b+c>0 \implies c > b-a $$
Given these conditions, the smallest values $a$, $b$ and $c$ can have are $a=1$, $b=c=3$. Unfortunately $x^2+3x+3$ doesn't have roots at all. Increasing $c$ can't help, so $b$ must be at least $4$, and so $c=4$ too.
$x^2+4x+4$ has a double root at $x_1=x_2=-2$ and if that is allowed by the conditions we have $abc=16$ and $a+b+c=9$.
If $a$ was larger, say $a=2$ conditions $(1)$ and $(2)$ would give $b\ge 5$ and $c\ge 4$, which would give $abc\ge40$, which is much larger so we can discount that.
If the double root is not allowed, we would need to go to higher $b$ still, so consider $b=5$ -- then we get $c=5$ too. Now at last there are two different roots (we don't need to compute them, only observe that the discriminant is positive). The product $abc$ is now $25$ which is still less than $40$, and the answer for $a+b+c$ is then $11$.