Winning strategy of Peter such that quadratic equations has Rational roots

Question

Peter picked three non zero real numbers $a,b,c$ and Alan arranges these numbers as coefficients of Quadratic equation $ax^2+bx+c=0$.

Peter wins if this Quadratic has distinct Rational roots, else Alan wins.

How can we prove that Peter always has winning strategy.

Since $a,b,c$ are non zero reals we have roots of quadratic as

$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Now Peter can win game only if $$b^2-4ac=k^2$$ where $k$ is Integer. But even though $b^2-4ac=k^2$ is integer, how can we say roots are rational?

Answer 1

It's enough to find one set of real numbers that Peter can choose to win. For example, $\{1, 2, -3\}$ will work. We can check that: \begin{align} x^2 + 2x - 3 &= (x+3)(x-1) \\ x^2 - 3x + 2 &= (x-2)(x-1) \\ 2x^2 + x - 3 &= (2x+3)(x-1) \\ 2x^2 - 3x + 1 &= (2x-1)(x-1) \\ -3x^2 + x + 2 &= -(3x+2)(x-1) \\ -3x^2 + 2x + 1 &= -(3x+1)(x-1) \end{align} and so both roots of all six polynomials Alan can form are rational and distinct.

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I found this by brute force, but in retrospect it seems obvious why this worked and how we can find lots more such triples. Since $1+2+(-3)=0$, $x=1$ will always be a root of the quadratic equation, and factoring out $x-1$ must produce a second rational root.

Answer 2

As pointed out by Misha Lavrov, ${1,2,-3}$ is a solution. But that's not the only case. In fact, Peter can choose any three integers $a$, $b$ and $c$ such that they sum up to zero i.e. $a+b+c=0$.

Proof: For the equation to have rational roots, the discriminant should be a perfect square. If $a+b+c=0$, then $b=-(a+c)$. Thus, the discriminant $D$ becomes $(a+c)^2-4ac=(a-c)^2$, which is clearly a perfect square. This method is general i.e. you can switch any of $a$, $b$ and $c$, always resulting in a perfect square discriminant.

Hope this helps :)