A nice way to solve this is to factor:
3\times 2^m = (n-1)(n+1)
Now clearly one factor on the right is a power of 2 since the factor 3 can only occur in one of them. Split it into cases depending on which it is and you should get the answers.
#4
robert2734
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0
of (n-1)(n+1) one factor has to be a power of 2 and the other 2*3=6. so the two possibiltites are 4*6 and 6*8.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?