I have seen this lemma elsewhere.
Suppose A is a domain, and f∈A[X]. Prove that
a−b∣f(a)−f(b)
I need to prove this.
f(a)−f(b)≡0(moda−b)
basically.
Let, a−b=c
f(a)−f(b)/(a−b)=f′(ξ)
for Some ξ∈(a,b).
But I dont see it showing divisibility
Answer
Let A be a domain with field of fractions K and let f(X)=a0+a1X+a2X2+⋯+anXn
be a polynomial in A[X]. Then for all a≠b∈A, f(b)−f(a)=(a0+a1b+⋯+anbn)−(a0+a1a+⋯+anan)=(b−a)(a1+a2b2−a2b−a+⋯+anbn−anb−a)
is certainly valid in K. But now note that both factors actually belong to A since b−a divides bk−ak for all k≥0.
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