TITLE: Prove that $\gcd(M, N)\times \mbox{lcm}(M, N) = M \times N$.
QUESTION [27 upvotes]: I'm not sure how to go about this proof. I just need help getting started. Is there a way to prove it algebraically?

REPLY [8 votes]: Please correct me if I'm wrong (I mean it!), but it seems to me that unique factorization (let alone the Euclidean algorithm or Bezout's identity) is not needed to prove this.  I'll give a proof of the generalization of the theorem for an arbitrary integral domain $R$ (but see notes right after the statement explaining how to specialize the theorem to the case $R = \mathbb{Z}$).

Let $R$ be an integral domain, and let $a, b$ be elements of $R$ having both a gcd $d$ and an lcm $m$ in $R$.  Then there exists a unit $u \in R$ such that $dm = uab$.

Note: as the wording suggests, uniqueness of gcds and/or lcms is not assumed here.  Also, when $R = \mathbb{Z}$, the only units (aka "invertible elements") are $\pm 1$, and furthermore, if one adopts the common convention that both the $\gcd(a, b)$ and the $\mathrm{lcm}(a, b)$ are defined to be positive, then the $\gcd(a, b)$ and $\mathrm{lcm}(a, b)$ are unique, and the theorem's punchline therefore reduces to $$\gcd(a, b) \times \mathrm{lcm}(a, b) = |ab|\;.$$

Proof:  Since $ab$ is a common multiple of $a$ and $b$, it follows from the definition1 of lcm that there exists a $\delta \in R$ such that $\delta m = ab$.  On the other hand, again by the definition of lcm, there exist $i,j\in R$ such that $ai = m = bj$.  Multiplying through by $\delta$ yields
$$\delta ai = \delta m = \delta bj\,.$$
Replacing the middle term with $ab\;(=\delta m)$ and cancelling $a$ from the left equality, and $b$ from the right one, we get that $\delta i = b$ and $\delta j = a$.  This shows that $\delta$ is a common divisor of both $a$ and $b$.
The definition of gcd now implies that there exists a $u \in R$ such that $\delta u = d$, and therefore,
$$dm = u\delta m = uab\;.$$
It remains to be shown that $u$ is a unit.  Now, again by the definition of gcd, there exist $r, s \in R$ such that $dr = a$ and $ds = b$.  Define $\mu = rds$.  Then
$$d\mu = dr\,ds = a\,ds = dr\,b = ab\;.$$
Cancelling $d$ in the first three equalities above shows that $\mu = as = br.$  This means that $\mu$ is a common multiple of both $a$ and $b$, and therefore, by the definition of lcm, there exists $v\in R$ such that $\mu = vm$.  
Substituting this expression for $\mu$ into the equality $d\mu = ab$ established above yields $d\,vm = vdm = ab$, which becomes $uvdm = uab$ after multiplying through by $u$.  Putting this together with the previously obtained $dm = uab$ gives $uvdm = dm$, which, after cancelling out the $dm$ factor reduces to $uv = 1$.  Hence $u$ is a unit.

1 This proof repeatedly invokes the definitions of gcd and lcm for a general integral domain; these definitions are entirely consistent with those for the gcd and lcm in $\mathbb{Z}$, but, for the sake of clarity and completeness, here they are: $d \in R$ is a divisor of $a \in R$ if there exists $r \in R$ such that $dr = a$; $d$ is a greatest common divisor (gcd) of $a, b\in R$ if (1) it is a divisor of both $a$ and $b$, and (2) any other divisor of $a$ and $b$ is also a divisor of $d$; note that, in general, for any $a, b \in R$ there may be any number of gcds—including none; if $d$ and $d^{\prime}$ are both gcds of $a$ and $b$, then they are associates (i.e. there exists a unit $u \in R$ such that $d = ud^{\prime}$); this is so because they must be divisors of each other.  Similarly, $m \in R$ is a multiple of $a\in R$ if there exists $q \in R$ such that $m = aq$; $m$ is a least common multiple (lcm) of $a, b\in R$ if (1) it is a multiple of both $a$ and $b$, and (2) any other multiple of $a$ and $b$ is also a multiple of $m$.  Similar considerations on uniqueness as those for gcds apply to lcms.  In particular, if $m$ and $m^{\prime}$ are both lcms of $a, b\in R$ then $m$ and $m^{\prime}$ are associates.