Mathematico | Mathematico

Problem 9 proof

Let $p, q \in N$ be distinct prime numbers.

Prove that

x^{2} + p x + q x + x + p q

cannot be factorised in the form

(x + α) (x + β)

where $α, β \in N$

Hint

Suppose, for contradiction, that it is possible. Then $α, β$ are positive integers such that both

\begin{aligned} α + β & = p + q + 1 \\ α β & = p q \end{aligned}

Hint

You should explain why both $α = p q, β = 1$ and $α = p, β = q$ are impossible choices. It is important that $p$ and $q$ are distinct and prime!

Solution

Suppose, for contradiction, that it is possible. Then $α, β$ are positive integers such that both

\begin{aligned} α + β & = p + q + 1 \\ α β & = p q \end{aligned}

Now, because $p$ and $q$ are prime numbers, ther are two cases to consider.

Case 1: $α = p, β = q$ . In this case

\begin{aligned} α + β & = p + q + 1 \\ p + q & = p + q + 1 \\ 0 & = 1 \end{aligned}

This is a clear contradiction.

Case 2: $α = p q, β = 1$ . In this case

\begin{aligned} α + β & = p + q + 1 \\ p q + 1 & = p + q + 1 \\ p q & = p + q \\ p q - p & = q \\ p (q - 1) & = q \end{aligned}

We have shown that $q = p (q - 1)$ . But $q$ is prime, so either $p$ or $q - 1$ must equal $1$ . However, $p$ is prime so cannot be $1$ , therefore $q - 1 = 1 \Rightarrow q = 2$ .

In this case, recall that we had

\begin{aligned} p q + 1 & = p + q + 1 \\ 2 p + 1 & = p + 3 \\ p & = 2 \end{aligned}

This is a contradiction because $p$ and $q$ are distinct, so cannot both equal $2$ .

We've considered all cases and each case results in a contradiction. We must conclude that the original hypothesis is incorrect, i.e. the expression cannot be factored as claimed.

Problem 9 proof ​

Problem 9 proof