There are 4 solutions meeting all of the requirements.
a^{2}=bd
ad=b^{2}c
Solving the first equation for d and substituting into the second gives
a*a^2/b = b^2*c
a^3 = c*b^3
c must be a perfect cube, but not 1 (since that would make a=b) so c=8.
From which it follows a=2b
Substitute into the first equation
(2b)^2=bd
4b=d
Since d<25, b<6.25 so we can list the 6 quadruplets
(a,b,c,d)
(2,1,8,4)
(4,2,8,8)
(6,3,8,12)
(8,4,8,16)
(10,5,8,20)
(12,6,8,24)
The solutions are bolded as the others do not have 4 distinct values.

Posted by Jer
on 20171228 10:13:39 