Wow, I interpreted this question in an entirely different way than everyone else.
The dogs of Oxford, I declare:
Numbered one third of a square.
I interpreted this to mean not a square number, but the four-sided polygon. We must assign a number to "a square," and then the number of dogs will be one third of that.
Let's say D is the number of dogs, and S is the number assigned to a square.
D = 1/3 * S
If one quarter left to roam,
Just a cube would stay at home.
A cube (C) is comprised of six square faces, so a cube is six squares.
C = 6 * S
This tells us that one fourth of the total number of dogs plus six times the number of dogs equals the total number of dogs:
D = 1/4 * D + 6 * D
At this point things are looking weird, so I started looking at answers and discovered that my interpretation was likely not the intended one, but let's run with it anyway. That last one simplifies down to:
D = 6.25D, and we all know there's exactly one solution to that:
D = 0.
That leaves the first two equations irrelevant unless one wants to know the values of the square and/or cube:
Both also zero.
What is the smallest possible number of dogs in Oxford?
Of course, the question implies that there should be multiple solutions, and in this case there is only the one. It's also a bit disappointing; there should be nonzero dogs in Oxford because dogs are awesome.
All of this points to my interpretation of the question being incorrect, but I thought my mistake was amusing and figured I'd share.
It might've helped if I had noticed the
mathematics tag. 😅