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Could any one give a basic algebra proof that for any polynomial P with integer coefficients, P(a)−P(b) is divisible by a−b.

3 Answers 3

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Hint $\rm\,\ mod\ a\!-\!b\!:\,\ a\equiv b\:\Rightarrow\: P(a)\equiv P(b),\ $ for all $\rm\:P\in \Bbb Z[x],\:$ by congruence sum, product rules.

Remark $\ $ For completeness here is a proof of the implication by induction on the degree of $\rm\,P.\:$ Clear if $\rm\:deg\ P = 0.\:$ Else $\rm\:P(x) = c + x\,Q(x)\:$ for $\rm\:c = P(0),\ Q(x)\in\Bbb Z[x]\:$ with $\rm\:deg\ Q < deg\ P.\:$ Therefore, by induction, $\rm\ a\equiv b\:$ $\Rightarrow$ $\rm\:Q(a)\equiv Q(b)\:$ $\Rightarrow$ $\rm\:P(a) = c + a\, Q(a)\equiv c + b\, Q(b)\equiv P(b),\ $ since, by the product and sum rules for congruences we have

$$\begin{eqnarray}\rm a&\equiv&\rm b,\ \ \ Q(a)&\equiv&\rm Q(b)\,\ &\Rightarrow&\rm\,\ a\, *\, Q(a)&\equiv&\rm b \,*\, Q(b)\\ \rm c &\equiv&\rm c,\ \ aQ(a)&\equiv&\rm bQ(b)\,\ &\Rightarrow&\rm\,\ c+aQ(a)&\equiv&\rm c+bQ(b) \end{eqnarray}$$

Notice that the effect of the induction is to lift congruence-preservation from the basic addition and multiplication operations to compositions of such, i.e. polynomial expressions. The innate structure will become clearer when you learn about polynomial rings and quotient (residue) rings.

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It is enough to show that the statement is true for monomials. $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots + ab^{n-2}+ b^{n-1})$$

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    Shouldn't we assue $a,b\in\Bbb Z$?2012-11-15
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    My equation holds for arbitrary numbers, though of course an *integer* factorization is only guaranteed if $a,b$ are integers. Therefore, the original divisibility question should require $a,b\in \mathbb Z$. Then again, $a-b$ divides $P(a)-P(b)$ as Polynomial in $\mathbb Z[a,b]$ if we consider $a,b$ indeterminates.2012-11-15
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    And what does $\Bbb Z[a,b]$ denote?2012-11-15
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    @PeterTamaroff The ring of polynomials in two indeterminates $a$ and $b$ with coefficients in the ring $\mathbb Z$ of integers.2012-11-15
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    Oh, OK. ${}{}{}{}$2012-11-15
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Let $Q(x) = P(a)-P(b+x)$ then $a-b$ is a root of $Q$ so $x - (a-b)|Q(x)$ and setting $x=0$ gives $a-b|P(a)-P(b)$.

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    This looks a bit weird, the point of this answer is we can reduce it to the lemma $a-x|Q(x)$ when a is a root of Q.2012-11-15