21we kno the lcm of any 2 nos. is the first no.*second no./gcd
lets denote gcd(m,n)=d
and m=ds for any natural no. s
and n=dt for any natural no. t
so,\frac{mn}{d}+d=m+n
so,\frac{ds\times dt}{d}+d=ds+dt
or,sdt+d=ds+dt
or,st+1=s+t
or,st-s+1-t=0
or,(s-1)(t-1)=0
so, either s=1 or t=1
therefore if s=1 m=d
or if t=1 n=d
if m=d m|n or if n=d n|m
so, one of the no. has to be divisible by another..........................
