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Prove that there is no choice of real numbers $p$ and $q$ such that

p^{2} = 2 (p + q - q^{2} - p q - 1)

Hint

Collect this into a quadratic in terms of $p$ . Get the discriminant in terms of $q$ .

Hint

When you have your expression for the discriminant, can you explain why it is negative for all values of $q$ ?

Solution

Let's get this in quadratic form in terms of $p$

\begin{aligned} p^{2} & = 2 (p + q - q^{2} - p q - 1) \\ p^{2} & = 2 p + 2 q - 2 q^{2} - 2 p q - 2 \\ p^{2} - 2 p - 2 q + 2 q^{2} + 2 p q + 2 & = 0 \\ p^{2} + 2 p (q - 1) + 2 (q^{2} - q + 1) & = 0 \end{aligned}

We have to prove that, no matter the choice of $q$ , this equation has no roots in $p$ .

Let's consider the discriminant. With

\begin{aligned} a & = 1 \\ b & = 2 (q + 1) \\ c & = 2 (q^{2} - q + 1) \end{aligned}

we get

\begin{aligned} Δ & = 4 (q - 1)^{2} - 4 (1) (2 (q^{2} - q + 1)) \\ = 4 (q^{2} - 2 q + 1 - 2 q^{2} + 2 q - 2) \\ = 4 (- 1 - q^{2}) \\ = - 4 (q^{2} + 1) \end{aligned}

Now, $q^{2} + 1 \geq 1 > 0$ so $- 4 (q^{2} + 1) < 0$ for every choice of $q \in R$ .

No matter the choice of $q$ , this equation has no solutions in $p$ .