MathLabs

Problem 3

Let mm and nn be arbitrary non-negative integers. Prove that (2m)!(2n)!m!n!(m+n)!\dfrac{(2m)!(2n)!}{m!n!(m+n)!} is an integer. (0!=10!=1).
Step 2 of 4: Compare neighboring values
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In plain words

Factorial ratios make neighboring terms differ by simple rational factors.

f(m,n−1)=f(m,n)m+n2(2n−1),f(m+1,n−1)=f(m,n)2m+12n−1f(m,n-1)=f(m,n)\dfrac{m+n}{2(2n-1)},\quad f(m+1,n-1)=f(m,n)\dfrac{2m+1}{2n-1}
Detailed analysis

For n≥1n\ge1, cancel factorials directly. The missing factors in the two neighboring expressions give the two displayed identities.