What you will read below is a brief exercise I completed in 2017 to resolve a point of curiosity in my mind.
As I recall, at the time I had just begun working on a side project: building a geometric model of stock price progressions to use as a base for discrete-probability option pricing. While the main project ultimately went nowhere, this output remains. Having had the proof below independently verified as correct, I now present it here for your reading pleasure. Enjoy!
Statement
Let N be a natural number with T factors (divisors), which is the product of V prime numbers, each of which is raised to some natural number A.
Then T is given by
Proof
The proof will be by induction.
Base Case
We begin with the base case: N is a power of a single prime number (i.e. V = 1). Then N = P1A1.
Let F = F(N) be the set of factors of N. Then F = {1, P1, … , P1A1-1, P1A1} and so has A1 + 1 elements.
Consider the additional case where V = 2. Then N = P1A1 P2A2.
P1A1 yields A1 + 1 distinct factors; P2A2 yields A2 + 1 distinct factors. Multiplying each relevant power of P1 by each relevant power of P2 yields the full set of factors of N. Therefore, N has (A1 + 1)(A2 + 1) factors.
Inductive Hypothesis
There exists some natural number W > 2 such that when N = P1A1…PW+1AW+1, N has (A1 + 1)(A2 + 1)…(AW + 1)(AW+1 + 1) factors.
Inductive Step
Assume our claim holds when V = W. Then there exists a natural number K = P1A1…PWAW with M = (A1 + 1)(A2 + 1)…(AW-1 + 1)(AW + 1) factors.
Let N = K * PW+1AW+1. Then N = P1A1…PW+1AW+1.
PW+1AW+1 yields (AW+1 + 1) distinct factors of N, each of which can be multiplied by the M distinct factors of K to yield all the distinct factors of N.
Therefore, every natural number N = P1A1…PVAV has (A1 + 1)(A2 + 1)…(AV-1 + 1)(AV + 1) distinct factors.
QED
Corollary 1
Let N = P1A1…PVAV be a natural number with D = |F(N)| = (A1 + 1)(A2 + 1)…(AV-1 + 1)(AV + 1) distinct factors. Then NX = (P1A1…PVAV)X = P1X*A1…PVX* AV has |F(NX)| = (X*A1 + 1)(X*A2 + 1)…(X*AV-1 + 1)(X*AV + 1) distinct factors.
Proof
QED
Corollary 2
For every natural number N = P1A1…PVAV with D = D(N|V) = (A1 + 1)(A2 + 1)…(AV-1 + 1)(AV + 1) distinct factors and for every natural number V,
Proof
N has V primes in its factorisation, each of which has an exponent of at least one in the factorisation.
D = (A1 + 1)(A2 + 1)…(AV-1 + 1)(AV + 1). Given that each exponent Ai is an integer greater than or equal to one [where i is an integer ranging from 1 to V],
QED
Corollary 3
Let D be some natural number M. If M is prime, then N = P1M-1.
Proof
The proof will be by contradiction.
M is a prime number, so it only has two factors: one and itself.
M can be written as M = (A1 + 1)(A2 + 1)…(AH-1 + 1)(AH + 1).
Suppose, without loss of generality, that M = (A1 + 1). Then this means that every other exponent Ai in the prime factorisation of M equals zero, which is impossible because every Ai must be a natural number, which by definition is positive.
Therefore, the only possible situation is that N has only one prime in its factorisation – P1 – and is consequently equal to P1M-1.
QED
Corollary 4
Let N be a natural number and G be its negative (i.e. G = –N). Then D(G|V) = |F(G)| = 2|F(N)| = 2D(N|V).
Proof
Suppose N has D factors.
G = -N, therefore every factor of N divides G.
However, because G is negative, negative one also divides G. Therefore, the negative of every factor of N is also a factor of G.
Therefore, |F(G)| = 2|F(N)| = 2D.
QED