This is really quite easy as p^2 can only be p times p or 1 times p^2.
Now, n^3 + 1 = (n + 1)(n^2 - n + 1)
So one possible solution is where one of the factors is equal to 1 and the other is equal to p^2. Or when they are both equal to each other. Test those out and you should find all the possible solutions for n.
#3
TenaliRaman
637
1
An alternative,
all odd p^2 is usually of the form
p^2 = 1 mod 4
n^3+1 = 1 mod 4
or n^3 = 0 mod 4
Now this is useful if and only if u have a list of primes ...
So first u can generate a list and then check for conditions ...
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 ?