From the 2018 Moroccan Mathematics Olympiad:
Let $(a,b) \in \mathbb{Z^2}$ such that $a+b$ is a solution of the equation $x^2+ax+b=0$. Find the smallest value of the product $ab$. ($\mathbb Z$ contains 0)
My Approach:
I've solved it by an extremely bashy method, which I don't think is worth showing. I just used the properties of quadratic equations, and basic divisibility properties. I then got all the possible solutions, and simply calculated the possible products. I found the m
Question :
Can one find the desired minimum without finding all solutions to the given equation?
Edit: Basically what I did, is find the equivalent equation: $(2a+b)^2=-b(b+2a+2)$, Now notice: $gcd(2a+b;b+2a+2)=1$ or $=2$, try all cases and get all solutions.
Edit 2: The solution given in the first edit, isn't the long bashy solution.
Best Answer
Starting from $2a^2+3ab+b^2+b=0$, use the quadratic formula to obtain an expression for $a$: $$a=\frac{-3b\pm \sqrt{9b^2-8(b^2+b)}}{4}$$ This only has integer solutions when $(b^2-8b)$ is a perfect square. This occurs when $b \in \{-1,0,8,9\}$
Next, an expression for $ab$ can be stated without finding any particular solutions to the original equation $$ab=\frac{-3b\pm \sqrt{9b^2-8(b^2+b)}}{4}\cdot b$$ Substituting the acceptable values of $b$ into that formula yields $-54$ as its smallest value, without ever identifying any of the values of $a$ or solutions to the original equation.