Doubt in understanding polynomial theorem $m-n \mid p(m)-p(n)$ Consider a polynomial $P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}$ with integer coefficients.
The difference $P(x)-$ $P(y)$ can be written in the form:-
$
a_{n}\left(x^{n}-y^{n}\right)+\cdots+a_{2}\left(x^{2}-y^{2}\right)+a_{1}(x-y)
$
in which all summands are multiples of polynomial $x-y .$
Hence, If $P$ is a polynomial with integer coefficients, then $P(a)-P(b)$ is divisible by $a-b$ for any distinct integers a and $b$
My doubt is that why this is valid for integer coefficients, why we cannot replace integer with rational or real coefficients ???
I know that $a$ is divisible by $b$ if $a=bc$ where $$c is integer but here we are talking about polynomials and I read that in polynomials $P$ is said to be divisible by $Q$ if $P=QB$ where $B$ is another polynomial.
You have to differ between divisibility of polynomials and divisibility in integers.
It is true that $x-y\mid p(x)-p(y)$ but this is not necessary true in integers. Example $$x-\sqrt{2}\mid x^2-2$$ but this is clearly not true in integers, say if $x=1$ then $1-\sqrt{2}$ does not divide $-1$ since it is not defined in $\mathbb{Z}$.
You have to differ between divisibility of polynomials and divisibility in integers.
It is true that $x-y\mid p(x)-p(y)$ but this is not necessary true in integers. Example $$x-\sqrt{2}\mid x^2-2$$ but this is clearly not true in integers, say if $x=1$ then $1-\sqrt{2}$ does not divide $-1$ since it is not defined in $\mathbb{Z}$.