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Spring 2016, problem 13

Consider the binomial coefficients (nk)=n!k!(nk)!, where k{1,2,,n1}. Determine all positive integers n for which (n1),(n2),,(nn1) are all even numbers.

Comments

Hubert
5 years ago

Hi, I'm Hubert from Paris ... Luca's theorem shows that n=2q;

using the Wikipedia article notations with p=2 we want N=(mn) even for 1nm1,

  • if m is not a power of 2, then mk=mj=1 for some j<k and the given formula says (m2j)(11)(mod2) with nj=1,nk=0,kj, odd;

  • now if m=2k,mj=0,j<k and n<m nk=0, and at least one of the n0,n1,nk1 is one, putting a 0=(01) in the given product, for each n(0,m), making each N even.

Note: this strange < is the symbol "less than", can someone fix that ?

I need your help now ... :-(

The browser interprets the < as the start of an html tag or something like that. Use \lt and \leq inside $ signs instead.

Nelix 5 years ago
Nelix
5 years ago

The number n has the required property, if and only if n=2k for some positive integer k.

I suspect, there's a nice proof for that, involving the binomial theorem and a little group theory, but all I could come up was this rather calculational Proof:

[The notation x:=floor(x) and x:=ceil(x) is used throughout.]

For every natural a, define ν(a):=b, where b is the largest integer, such that 2b divides a. Or put another way: ν(a) is the numbert of twos in the prime factorization of a.

A particular case of a result known as Legendre's formula states, that for all aN:

ν(a!)=i=1a2i

Also, from the definition of the binomial coefficients, we have:

ν((nk))=ν(n!)ν(k!)ν([nk]!) (*)

Now assume, that n is not a power of 2. So n lies strictly between two powers 2j and 2j+1 of two, i.e. there are j,rN, with 0<r<2j, such that n=2j+r.

Then ν(n!)=ν([2j+r]!)=i=12j+r2i=ji=12ji+r2i=ν([2j]!)+ν(r!). Note, that the second manipulation only works, because we assume, that r<2j. Plugging this into the equation (*), we get:

ν((n2j))=ν((2j+r2j))=ν([2j]!)+ν(r!)ν([2j]!)ν(r!)=0

Therefore (n2j) is odd so that n does not have the property required in the problem statement.

This concludes the first half of the proof. We now assume, that n=2j for some jN and prove it does indeed have said property.

Pick a kN, such that 0<k<2j. Then:

ν((nk))=ν((2jk))=ν([2j]!)ν([2jk]!)ν(k!)=ji=1k2ik2i. By the definition of the rounding functions, none of the summands is negative and - from 0<k<2j - the last one is equal to 1. Therefore ν((nk))>=1 and so (nk) is even. This completes the proof.

As an aside, using the general form of Legendre's formula, the proof takes little modification, to show that:

A prime number p divides all of (n1),(n2),,(nn1) iff n is a power of p.