Note that and are natural numbers and you have to find all the possibilities. I only found that or and .
first, a symetry exists among a, b, c, d
so, suppose a=<b=<c=<d
in this way, 4/a>=1
a>=1
well, you can suppose a=1 or a=2 or a=3 or a=4
* if a=1
there is no solutions
* if a=2
then,
or b=<c=<d
so 3/b>=1/2
b=<6
in this way, b=1 or b=2 or b=3 or b=4 or b=5 or b=6
**if b=1 : no solutions
**if b=2 : no solutions
**if b=3 :
or c=<d
so 2/c=>1/6
c<=12
in this way, c=1 or c=2 or c=3 or c=4 or c=5 or c=6 or c=7 or c=8 or c=9 or c=10 or c=11 or c=12
***if c=1 or c=2 or c=3 or c=4 or c=5 or c=6 : no solutions
***if c=7:
so, d=42
solution 1 : a=2, b=3, c=7, d=42
***if c=8 :
so, d=24
solution 2 : a=2, b=3, c=8, d=24
***if c=9 :
so, d=18
solution 3 : a=2, b=3, c=9, d=18
***if c=10 :
so, d=15
solution 4 : a=2, b=3, c=10, d=15
***if c=11 : no solution
***if c=12 :
so, d=12
solution 5 : a=2, b=3, c=12, d=12
**if b=4:
.....
it seems too long....but for sure it's an exhaustive way to solve this problem.