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Theorem: binomial(n^2, n)/(n+1) is an integer for n >= 2.
Proof 1 from William J. Keith, May 08 2010:
binomial(n^2, n) * 1/(n+1)
= (n^2)(n^2-1)(n^2-2)!/((n^2-n)!n(n-1)(n-2)!) * 1/(n+1)
= n (n^2-2)!/((n^2-n)!(n-2)!) = n * binomial(n^2-2,n-2). QED
Proof 2 from Max Alekseyev, May 08 2010:
Recall that the valuation of m! w.r.t. prime p equals the sum floor(m/p^i) over i=1,2,3,...
Moreover, if m=a+b where a and b are nonnegative integers, then floor(m/p^i) - floor(a/p^i) - floor(b/p^i) >= 0.
Let n>1. To prove that binomial(n^2, n)/(n+1) is an integer, it is enough to show that its valuation w.r.t. any prime p is nonnegative.
It is clear that trouble may come only from primes dividing n+1.
Let valuation(n+1,p)=k > 0, i.e., n+1=p^k*m where prime p does not divide m.
Then n = p^k*m - 1, n^2 = p^(2k)*m^2 - 2*p^k*m + 1 and n^2 - n = p^(2k)*m^2 - 3*p^k*m + 2.
It is easy to check that floor(n^2/p^i) - floor(n/p^i) - floor((n^2-n)/p^i) = 1 for i=1,2,...,k if p>2 and for i=2,3,...,k+1 if p=2, implying that valuation(binomial(n^2, n)/(n+1),p) >= 0. QED
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