Factoring Combinatorial Functions

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Homework Statement

Define $${x \choose n}=\frac{x(x-1)(x-2)...(x-n+1)}{n!}$$ for positive integer n. For what values of positive integers n and m is $$g(x)={{{x+1} \choose n} \choose {m}}-{{{x} \choose n} \choose {m}}$$ a factor of $$f(x)={{{x+1} \choose n} \choose {m}}$$?

Homework Equations

The idea that the roots of a polynomial must be roots of its factors perhaps.

The Attempt at a Solution

I quickly found by brute force that m=n=1 and m=n=2 worked, but I’m not sure how to get a general result. Any help would be appreciated.

 

[.Scott]

Define $${x \choose n}=\frac{x(x-1)(x-2)...(x-n+1)}{n!}$$ for positive integer n. For what values of positive integers n and m is $$g(x)={{{x+1} \choose n} \choose {m}}-{{{x} \choose n} \choose {m}}$$ a factor of $$f(x)={{{x+1} \choose n} \choose {m}}$$?

I would start by finding $A$ for: ${x \choose n} = A{(x+1) \choose n}$.

Then $g(x)$ becomes ${{{x+1} \choose n} \choose {m}} - {A{{x+1} \choose n} \choose {m}}$

Then find $B$ for: ${{{x+1} \choose n} \choose {m}} = B {A{{x+1} \choose n} \choose {m}}$

Then $g(x)$ becomes $f(x)(1-B)$

 

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I would start by finding $A$ for: ${x \choose n} = A{(x+1) \choose n}$.

Then $g(x)$ becomes ${{{x+1} \choose n} \choose {m}} - {A{{x+1} \choose n} \choose {m}}$

Then find $B$ for: ${{{x+1} \choose n} \choose {m}} = B {A{{x+1} \choose n} \choose {m}}$

Then $g(x)$ becomes $f(x)(1-B)$

I find $A=1-\frac{n}{x+1}$. Also, I believe $g(x)=(1-\frac{1}{B})f(x)$. $B$ is rather nasty, and I’m not sure what to do with it.