just filling in some gaps in proofs of dm=ab.

d=(a,b)=gcd(a,b)
m=[a,b]=lcm(a,b)


Meserve, Dover
a=p1d, b=p2d, (p1,p2)=1…..definition of d
p2a=p1b=p1p2d = c, a common multiple.
Assume c=rm…. (missing in Meserve)
m=q1a=q2b
p2a=rq1a → p2=rq1
p1b=rq2b → p1=rq2
r=1 because (p1,p2)=1 → c=m
p1p2d=m
dp1p2d=dm
dm=ab

Birkhoff MacLean
d=sa+tb, not obvious, see BM
a=p1d, b=p2d,
m=q1a=q2b, (q1,q2)=1…..definition of m
md=saq2b+tbq1a
md=(sq2+tq1)ab
md=kab
q1ad=kap2d → q1=kp2
q2bd=kp1db → q2=kp1
k=1 because (q1,q2)=1
dm=ab…..(not proved in BM)

google “gcd(a,b) x lcm(a,b)” for a proof using primes.