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 4 of 4: Induct from the base row
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In plain words

The integral central binomial row propagates upward through an integer recurrence.

f(m,0)=(2mm)∈Zf(m,0)=\binom{2m}{m}\in\mathbb Z
Detailed analysis

For n=0n=0, f(m,0)=(2mm)f(m,0)=\binom{2m}{m} is an integer. If all values with second argument n−1n-1 are integral, the recurrence makes f(m,n)f(m,n) integral. Induction on nn proves the claim.