gcd((n−1k−1),(nk+1),(n+1k))=gcd((n+1k+1),(nk−1),(n−1k))\gcd({n-1\choose k-1},{n \choose k+1},{n+1 \choose k})=\gcd({n+1\choose k+1},{n\choose k-1},{n-1\choose k})gcd((k−1n−1),(k+1n),(kn+1))=gcd((k+1n+1),(k−1n),(kn−1))
在网上找了好久都没找到证明
