proof $z \mid b$ and $w\mid b$

Question

Question I'm working on: Let $a,b$ be integers with $b$ not equal to $0$. suppose $x^2+ax+b=0$ and $x=z,w$. If $z,w$ are integers, show that $z\mid b$ and $w\mid b$.

Is it sufficient for me to show that $x^2+ax+b=(x-z)(x-w)$ so $b=zw$?

Answer 1

Since $w$ and $z$ are solution of the equation $$x^2+ax+b=0$$ we have $$w^{2}+aw+b=0 \Rightarrow b=-w(w+a) \Rightarrow w|b$$

In the same way for $z$.

Regards!!

Answer 2

Here is a better approach:

We need to show $z|b$ and $w|b$. What does it mean for $z$ to "divide" $b$? It means there is some integer $k$ such that $z \cdot k = b$. Well, what are we given? We are given that if $x = z$ or $x = w$, then $x$ satisfies $x^{2} + ax + b = 0$.

But that means $z^{2} + az + b = 0$, or $z^{2} + az = -b$. But then that means $z(z + a) = - b$. Since $z + a$ is an integer (because $z$ and $a$ are integers), then we have found an integer $k$ such that $z \cdot k = -b$, so $z$ divides $-b$. Is it clear that if $z$ divides $-b$, then $z$ divides $b$?

You can use the same argument for $w$. Does this argument make sense to you? Do you understand every step?