$f(x)$ is a polynomial with integer coefficients. show that $$\frac{f(b)-f(a)}{b-a}$$ is always an integer for all distinct integers $a,b$

Question

freckles · Answer

say 
$$f(x)=m_0+m_1x+m_2x^2+ \cdots + m_nx^n \ \text{ so we want \to show there exist integer } k \text{ such that } \ (b-a)|[(m_0+m_1b+m_2b^2+\cdots m_nb^n)-(m_0+m_1a+m_2a^2+ \cdots +m_nb^n)] \ \text{ so we have we want \to show } \ (b-a)|[m_1(b-a)+m_2(b^2-a^2)+m(b^3-a^3)+\cdots m_n(b^n-a^n)]$$

but all of those will have a factor (b-a) in common

kainui · Answer

$$f(a) \equiv f(b) \mod b-a$$ substitute $k=b-a$
$$f(a) \equiv f(k+a) \mod k$$
Since f(k+a) is just a bunch of integers multiplying binomials, all the higher terms containing k in the expansion cancel out and all you're left with is the f(a) terms. Done!

freckles · Answer

$$ (b-a) \ (b^2-a^2)=(b-a)(b+a) \ (b^3-a^3)=(b-a)(b^2+ab+a^2) \  \cdots \ (b^n-a^n)=(b-a)(b^{n-1}+ab^{n-2}+ \cdots +a^{n-1})


$$

freckles · Answer

$$m_1(b-a)+m_2(b^2-a^2)+\cdots m_n(b^n-a^n) \ =(b-a)(m_1+m_2(b+a)+\cdots +m_n(b^{n-1}+ab^{n-2}+ \cdots +a^{n-1}))$$

freckles · Answer

where that big ugly thing next to the (b-a) is an integer which i called k above

kainui · Answer

Ahhh yeah that's a good too! There's something kind of fascinating about this, that you can always factor out a $b-a$ out of $b^n-a^n$ but $b^n+a^n$ will never be $c^n$ hahaha.

kainui · Answer

(for n>2 of course haha)

rational · Answer

Both are very neat xD Was stuck at this for a while while going through the solution for a related problem http://math.stackexchange.com/questions/83592/prove-that-fx-has-no-integer-roots

kainui · Answer

Oh I don't know what does this mean $$\mathbb{Z}[x]$$