 26
 1
 Problem Statement
 Let a and b be positive integers, and let m be an integer such that ab = m (a,b). Without using the prime factorization theorem, prove that (a,b) [a,b] = ab by verifying that m satisfies the definition of [a,b], as follows:
 Relevant Equations

The notation [a,b] is used for the least common multiple of a and b, where
i: [a,b] is a multiple of both a and b.
ii: Any multiple of both a and b is a multiple of m.
The notation (a, b) is used for the greatest common divisor of a and b.
First, prove that [a,b] is a divisor of both a, b:
By definition, (a,b)  a and (a,b)  b.
So for some integers j and k,
a = (a,b) j and b = (a,b) k.
Multiplying the last two equations yields: ab = (a,b)^2 * jk.
But we are given that ab = m (a,b).
Equating the two expression for ab yields: (a,b)^2 * jk = m (a,b).
Dividing both sides by (a,b) gives: (a,b) jk = m.
Rearranging the factors two different ways:
[(a,b) j ] k = m and [(a,b) k] j = m.
But [(a,b) j ] k = a and [(a,b) k ] j = b.
So
ak = m and bj = m.
But this means that am and bm. So m is a common multiple of a and b. So, I proved the first half of the definition.
Now I want to prove the second half of the definition, namely that any other multiple of a and b is a multiple of m. It seems like it should be trivial, but I can’t seem to come up with the logic. Of course, I can come up with many examples, but no “reason” why it must be so.
By definition, (a,b)  a and (a,b)  b.
So for some integers j and k,
a = (a,b) j and b = (a,b) k.
Multiplying the last two equations yields: ab = (a,b)^2 * jk.
But we are given that ab = m (a,b).
Equating the two expression for ab yields: (a,b)^2 * jk = m (a,b).
Dividing both sides by (a,b) gives: (a,b) jk = m.
Rearranging the factors two different ways:
[(a,b) j ] k = m and [(a,b) k] j = m.
But [(a,b) j ] k = a and [(a,b) k ] j = b.
So
ak = m and bj = m.
But this means that am and bm. So m is a common multiple of a and b. So, I proved the first half of the definition.
Now I want to prove the second half of the definition, namely that any other multiple of a and b is a multiple of m. It seems like it should be trivial, but I can’t seem to come up with the logic. Of course, I can come up with many examples, but no “reason” why it must be so.