Combinatorial Function Factoring
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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 fairly quickly realized that m=n=1 and m=n=2 both worked, and thus perhaps m=n in general will work, though I’m not sure. The quotient upon division must obviously be linear. I suppose we’d have to ensure that all roots of $f$ are also roots of $g$, but I’m not sure about how to do so. Any help would be appreciated
After a bit more thought, I realized it’s equivalent to having consecutive roots differing by one for $f(x)$, but I still don’t know how to proceed from there.
combinatorics algebra-precalculus polynomials factoring
asked 2 days ago
Don Jon
133
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up vote
2
down vote
favorite
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 fairly quickly realized that m=n=1 and m=n=2 both worked, and thus perhaps m=n in general will work, though I’m not sure. The quotient upon division must obviously be linear. I suppose we’d have to ensure that all roots of $f$ are also roots of $g$, but I’m not sure about how to do so. Any help would be appreciated
After a bit more thought, I realized it’s equivalent to having consecutive roots differing by one for $f(x)$, but I still don’t know how to proceed from there.
combinatorics algebra-precalculus polynomials factoring
asked 2 days ago
Don Jon
133
Tedious CAS computations show that $g(x)$ is not a factor of $f(x)$ for $m=n=3$. Nor is it when $left{m,nright} = left{2,3right}$.
– darij grinberg
2 days ago
Are you sure $m=n=2$ works? If we set x=9, I get ${{10 choose 2} choose 2} - {{9 choose 2} choose 2} = 360$, while ${{10 choose 2} choose 2} = 990$, which doesn't have 360 as a factor.
– Todor Markov
16 hours ago
Yes— the quotient is just $frac{x+2}{4}$, which won’t always give an integer for a fixed $x$.
– Don Jon
11 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 fairly quickly realized that m=n=1 and m=n=2 both worked, and thus perhaps m=n in general will work, though I’m not sure. The quotient upon division must obviously be linear. I suppose we’d have to ensure that all roots of $f$ are also roots of $g$, but I’m not sure about how to do so. Any help would be appreciated
After a bit more thought, I realized it’s equivalent to having consecutive roots differing by one for $f(x)$, but I still don’t know how to proceed from there.
combinatorics algebra-precalculus polynomials factoring
asked 2 days ago
Don Jon
133
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 fairly quickly realized that m=n=1 and m=n=2 both worked, and thus perhaps m=n in general will work, though I’m not sure. The quotient upon division must obviously be linear. I suppose we’d have to ensure that all roots of $f$ are also roots of $g$, but I’m not sure about how to do so. Any help would be appreciated
After a bit more thought, I realized it’s equivalent to having consecutive roots differing by one for $f(x)$, but I still don’t know how to proceed from there.
combinatorics algebra-precalculus polynomials factoring
combinatorics algebra-precalculus polynomials factoring
asked 2 days ago
Don Jon
133
asked 2 days ago
Don Jon
133
edited yesterday
asked 2 days ago
Don Jon
133
asked 2 days ago
Don Jon
133
asked 2 days ago
Don Jon
133
133
Tedious CAS computations show that $g(x)$ is not a factor of $f(x)$ for $m=n=3$. Nor is it when $left{m,nright} = left{2,3right}$.
– darij grinberg
2 days ago
Are you sure $m=n=2$ works? If we set x=9, I get ${{10 choose 2} choose 2} - {{9 choose 2} choose 2} = 360$, while ${{10 choose 2} choose 2} = 990$, which doesn't have 360 as a factor.
– Todor Markov
16 hours ago
Yes— the quotient is just $frac{x+2}{4}$, which won’t always give an integer for a fixed $x$.
– Don Jon
11 hours ago
add a comment |
Tedious CAS computations show that $g(x)$ is not a factor of $f(x)$ for $m=n=3$. Nor is it when $left{m,nright} = left{2,3right}$.
– darij grinberg
2 days ago
Are you sure $m=n=2$ works? If we set x=9, I get ${{10 choose 2} choose 2} - {{9 choose 2} choose 2} = 360$, while ${{10 choose 2} choose 2} = 990$, which doesn't have 360 as a factor.
– Todor Markov
16 hours ago
Yes— the quotient is just $frac{x+2}{4}$, which won’t always give an integer for a fixed $x$.
– Don Jon
11 hours ago
Tedious CAS computations show that $g(x)$ is not a factor of $f(x)$ for $m=n=3$. Nor is it when $left{m,nright} = left{2,3right}$.
– darij grinberg
2 days ago
Tedious CAS computations show that $g(x)$ is not a factor of $f(x)$ for $m=n=3$. Nor is it when $left{m,nright} = left{2,3right}$.
– darij grinberg
2 days ago
Are you sure $m=n=2$ works? If we set x=9, I get ${{10 choose 2} choose 2} - {{9 choose 2} choose 2} = 360$, while ${{10 choose 2} choose 2} = 990$, which doesn't have 360 as a factor.
– Todor Markov
16 hours ago
Are you sure $m=n=2$ works? If we set x=9, I get ${{10 choose 2} choose 2} - {{9 choose 2} choose 2} = 360$, while ${{10 choose 2} choose 2} = 990$, which doesn't have 360 as a factor.
– Todor Markov
16 hours ago
Yes— the quotient is just $frac{x+2}{4}$, which won’t always give an integer for a fixed $x$.
– Don Jon
11 hours ago
Yes— the quotient is just $frac{x+2}{4}$, which won’t always give an integer for a fixed $x$.
– Don Jon
11 hours ago
add a comment |
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Tedious CAS computations show that $g(x)$ is not a factor of $f(x)$ for $m=n=3$. Nor is it when $left{m,nright} = left{2,3right}$.
– darij grinberg
2 days ago
Are you sure $m=n=2$ works? If we set x=9, I get ${{10 choose 2} choose 2} - {{9 choose 2} choose 2} = 360$, while ${{10 choose 2} choose 2} = 990$, which doesn't have 360 as a factor.
– Todor Markov
16 hours ago
Yes— the quotient is just $frac{x+2}{4}$, which won’t always give an integer for a fixed $x$.
– Don Jon
11 hours ago