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.
Proof of an Integer polynomial property
3 Answers
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.
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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0Oh, OK. ${}{}{}{}$ – 2012-11-15
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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0This 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