# About factoring trinomials over $\mathbb{Z}$

We were taught in school an algorithm to factor a trinomial of the form

$$x^2\pm bx\pm c$$

with $b,c\in \mathbb{Z}^+$. Assuming the best scenario (that the polynomial has both roots in $\mathbb{Z}$), the algorithm is described as follows:

The factorization has the form $(x\ @\ \alpha)(x\bigtriangleup \beta)$ where $@$ is the second sign of the trinomial (the sign of the coefficient of $x$) and $\bigtriangleup$ is the multiplication of the signs of the coefficient of $x$ and the constant term and $\alpha, \beta\in \mathbb{Z}^+$, $\alpha\geq\beta$.

Here can be noted that $\alpha$ and $\beta$ must satisfy $\alpha\beta = c$ and $\alpha - \beta=x$ if $@\cdot \bigtriangleup=-1$ or $\alpha + \beta=x$ if $@\cdot \bigtriangleup=1$, moreover, this conditions are SUFFICIENT to produce the factorization.

This allows us to obtain the factorization by finding two positive integers $\alpha$ and $\beta$ that satisfy those relations (which depends entirely on the polynomial given in the form $x^2\pm bx\pm c$), my question is: Why are those $\alpha$ and $\beta$ always unique?, I'll explain myself with an example.

If we want to factor $x^2+x-6$, we argue as follows: The factorization has the form $(x+\alpha)(x-\beta)$ where $\alpha\beta=6$ and $\alpha - \beta = 1$, "two numbers whose difference is $1$ and product is $6$... $3$ and $2$!" (school reasoning) so the factorization is $(x+3)(x-2)$. Since $\mathbb{Z}[x]$ is a UFD, the solution $(\alpha, \beta)$ must be unique, but this seems not to be trivial if we look at the equations system $$x - y = b, \ \ xy=c\ .$$ Can you give a proof of the fact that if an equations system of the form $$x - y = b, \ \ xy=c\ \mbox{ or } x + y = b, \ \ xy=c$$ with $b,c\in \mathbb{Z}^+$ has a solution $(\alpha,\beta)$ with $\alpha,\beta\in\mathbb{Z}^+$ then the solution is unique?