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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.

Martin Sleziak
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Ishan
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    Yes, it is also true in the sense of polynomial divisibility for any coefficients. – Michal Adamaszek May 12 '20 at 09:06
  • @MichalAdamaszek really, so why they written integer coefficients ? – Ishan May 12 '20 at 09:10
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    Because the problem they want you to solve is about one integer dividing another, not about one polynomial dividing another (although the solution method is exactly the same). And that requires integers to even make sense. For example the problem about integer divisibility can be solved by someone who has no clue about divisibility between 2-variable polynomials. – Michal Adamaszek May 12 '20 at 09:12
  • The Factor Theorem works for polynomials with coef's in any commutative ring $R.,$ When we evaluate the polynomial divisibility at $,x,y\in R,$ we obtain a divisibility in the coef ring $R,$ (vs. $\Bbb Z),,$ e.g. consider the coef ring $R = \Bbb Z + \Bbb Z\ i$ = gaussian integers. – Bill Dubuque May 14 '20 at 18:50

1 Answers1

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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 $\mathbb{Z}$.

Martin Sleziak
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nonuser
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  • But $,1!-!\sqrt 2, \mid, {-}1,$ is a valid divisibility in the coef ring $,\Bbb Z[\sqrt 2].,$ The polynomial divisibility always specializes to a valid divisibility of the coef ring. This is a special case of the universality of polynomials. – Bill Dubuque May 14 '20 at 18:56