Winning strategy of Peter such that quadratic equations has Rational roots

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?

• Just a minor change. The quadratic looks more familiar if we use the first three letters of the alphabet. – Tito Piezas III Apr 4 '17 at 2:54
• why do you think peter does have a winning strategy? – fleablood Apr 4 '17 at 3:03
• its given in the book that Peter always has winning strategy – Umesh shankar Apr 4 '17 at 3:05
• It might be. But peter must chose them so that then largest absolute middle value are different parity. – fleablood Apr 4 '17 at 3:07
• ya i also thought of it thats why i have posted here – Umesh shankar Apr 4 '17 at 3:07

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.
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.
• No, I had Mathematica guess for me. I tried a bunch of triples of integers checking if each of $a^2-4bc$, $b^2-4ac$, and $c^2-4ab$ were integers, and found this as the simplest. – Misha Lavrov Apr 4 '17 at 3:18