Originally Posted by

**jarman007** Standard form of quadratic equations is ax^2+bx+c=0

if a=1 , b and c are integers and the roots are rational numbers

then prove that roots must be integers?

This, in fact, is true of a polynomial of arbitrary degree; it was proven first by Gauss, in this much more general form :

Let $\displaystyle f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ be a monic polynomial with integer coefficients.

Suppose $\displaystyle f(x)=p(x)q(x)$ with $\displaystyle p(x),q(x) \in \mathbb{Q}[x]$. Then in fact $\displaystyle p(x),q(x) \in \mathbb{Z}[x]$.

The proof is quite simple using algebra. Suppose $\displaystyle p(x),q(x) \notin \mathbb{Z}[x]$; then there is a least positive integer $\displaystyle m$ and a least positive integer $\displaystyle n$ such that $\displaystyle mp(x)$ and $\displaystyle np(x)$ are in $\displaystyle \mathbb{Z}[x]$. It's easy to see that there is no integer $\displaystyle >1$ which divides all the coefficients of $\displaystyle mp(x)$ (or else $\displaystyle m$ could be made smaller), and similarily for $\displaystyle np(x)$.

Take any prime $\displaystyle p$ which divides the product $\displaystyle mn$, and let $\displaystyle h(x) \mapsto \overline{h(x)}$ denote the canonical projection $\displaystyle \mathbb{Z}[x] \longrightarrow (\mathbb{Z}/p\mathbb{Z})[x]$. Then $\displaystyle 0 = \overline{mnf(x)} = \overline{mnp(x)q(x)} = \overline{mp(x)}\ \overline{nq(x)}$. Since $\displaystyle (\mathbb{Z}/p\mathbb{Z})[x]$ is an integral domain, we must have $\displaystyle \overline{mp(x)} = 0$ or $\displaystyle \overline{nq(x)} = 0$. But $\displaystyle \overline{mp(x)} \neq 0$ and $\displaystyle \overline{nq(x)} \neq 0$ or else all the coefficients of $\displaystyle mp(x)$ or all the coefficients of $\displaystyle nq(x)$ would be divisible by $\displaystyle p$, which contradicts the choice of $\displaystyle m,n$, as described above. $\displaystyle \square$