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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.

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up vote 1 down vote accepted

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. – sperners lemma Nov 15 '12 at 19:05

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 $a,b\in\Bbb Z$? – Pedro Tamaroff Nov 15 '12 at 19:08
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. – Hagen von Eitzen Nov 15 '12 at 19:13
And what does $\Bbb Z[a,b]$ denote? – Pedro Tamaroff Nov 15 '12 at 19:15
@PeterTamaroff The ring of polynomials in two indeterminates $a$ and $b$ with coefficients in the ring $\mathbb Z$ of integers. – Hagen von Eitzen Nov 15 '12 at 21:00
Oh, OK. ${}{}{}{}$ – Pedro Tamaroff Nov 15 '12 at 22:02

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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